Transcript for:
2D Projectile Motion Overview

hey everyone in this video we're going to learn about two-dimensional projectile motion we'll start with an overview of 2D projectile motion and look at the equations that we're going to use we'll spend some time exploring two different examples an object launched horizontally from an initial height and an object launched from the ground at an angle in each example we'll use the equations to find the X and Y positions and see how they change over time then we'll do the same thing for the X and Y velocities we'll also talk about the motion graphs for each example then we'll learn more about the range of projectile motion and how it relates to the launch angle finally we'll check out the graphs for some other examples at the end we'll do a summary of the main points let's start with an overview of two-dimensional projectile motion we covered the basics of projectile motion in the introduction video so watch that one first in the last video we learned about one-dimensional projectile motion where an object only moves up and down 2D projectile motion just add add a horizontal velocity and it's very similar to 1D projectile motion so watch that video if you haven't already in 1D projectile motion we only dealt with Motion in the y direction in 2D projectile motion we have Motion in the X and Y directions whenever an object moves sideways through the air it's in 2D projectile motion this shows up everywhere ESP especially in sports the initial velocity and angle determine how far the ball travels and how high it is at each point they also determine the direction of the Velocity at every point along the path changing the initial angle just a few degrees changes the entire trajectory even if something moves really fast gravity still causes it to fall down so its path is never a straight line we usually focus on the path of one object but 2D projectile motion can help us understand other things like the shape and range of a stream of water if we ignore air resistance an object always falls at the same rate due to gravity 9.8 m/s squared so a ball that's launched horizontally will hit the ground at the same time as a ball that falls straight down so what's different about 2D projectile motion that we haven't already covered in 1D projectile motion an object moves in the y direction and it doesn't move sideways the initial velocity is also the initial y velocity the Velocity in the X direction is zero but in 2D projectile motion an object does move sideways so it has some velocity in the X Direction the initial velocity is not vertical it's usually at an angle so now we have an initial y velocity and an initial x velocity the acceleration in the y direction is always 9.8 m/s squ downwards but there's no acceleration in the X Direction ax is zero so the horizontal velocity VX X is constant the entire time the final x velocity at any time is the same as the initial x velocity in 2D projectile motion we call the horizontal displacement the range of the projectile motion that's how far the object travels in the X direction for 2D projectile motion we're going to use the same kinematic equations for the Y Direction but we we also need some equations in the X Direction since there's no horizontal acceleration we won't use the X equations with acceleration so we just have the displacement and velocity equations and remember the equations can be rearranged in different ways if we say up is the positive y direction then we're going to plug in negative G NE 9.8 m/s squ for a y finally even though the variables and the X and Y equations are different time shows up in both directions T has the same value in both equations at the same moment in time so we can use time to connect the X and Y equations in 1D projectile motion there's only three initial conditions in 2D projectile motion we have some new scenarios let's say up is the positive y direction and right is the positive X direction if the initial angle is above the horizontal the initial y velocity points upwards so it's positive the final height could be the same as the initial height like if it starts and ends at the ground the final height could be greater than the initial height like if it's going over a wall or landing at a higher point or the final height could be less than the initial height like if it's launched from a cliff if that's the case the initial velocity could also be horizontal so the initial y velocity is zero or the angle could even be below the horizontal so the initial y velocity points downwards and it's negative the only difference between these scenarios is the initial and final values and the positive and negative signs but the motion behaves the same way in each case we're going to use the same equations and solve problems the same way for each one so that's an overview of 2D projectile motion now let's explore some examples example one is an object with an initial horizontal velocity a ball starts at a height of 19.6 M with a horizontal velocity of 10 m/s for 2D projectile motion here's some questions we might see how far does the ball travel how long is the ball in the air and what is the speed of the ball when it lands these questions are just asking for a value at a specific point in the motion like when the object hits the ground but let's start by exploring this motion and come back to these questions later with projectile motion it's not really important why why the object has its initial height and velocity in this example the ball is shot out of a cannon on a cliff but we really just care about the projectile motion of the ball so all we need is the initial height and the initial velocity now that this is two-dimensional motion we need to choose the origin where X and Y are zero and which directions are positive you can set this up differently but let's place the origin at the gra ground directly below the initial position and we'll say right is the positive X Direction and up is the positive y direction using this coordinate system the ball's initial X position is 0 m and the initial y position is 19.6 M we could also write the initial position using coordinates as a side note we're usually describing the projectile motion of an object's Center of mass when we draw a picture the object is going to have some width and height so keep in mind that we're describing the center of the ball so it hits the ground when its Center has a y position of zero the initial velocity is 10 m/s and it's in the horizontal direction that means the initial x velocity is 10 m/s the initial y velocity is zero because this Vector has has no y component it only points in the X Direction since this is projectile motion the acceleration due to gravity G is 9.8 m/s squar downwards since we chose up to be the positive y direction then a y the vertical acceleration would be netive G 9.8 m/s squ so that's our setup now let's see what this motion looks like the ball moves to the right and falls down at the same time what's happening to the X position over time what about the Y position and what do you think is happening to the Velocity in the horizontal and vertical directions are they increasing decreasing or staying the same here's some grid lines and snapshots of the ball's position at half second intervals how much does the ball move to the right during each interval and how much does it move down if we draw lines from the ball to the Y AIS we can see how the Y position is changing over time the spacing between each line gets longer as the ball moves it falls a greater distance in the vertical Direction during each second or each half second now let's look at the the motion in the X Direction the spacing between each snapshot is the same the ball moves the same horizontal distance during each interval so why are the vertical and horizontal motions different 2D projectile motion is one example of an important Concept in physics that we mentioned in a previous lesson an object's X motion and Y motion are independent of each other the vertical MO motion does not affect the horizontal motion and the horizontal motion doesn't affect the vertical motion we know that objects in projectile motion experience a downwards acceleration due to gravity since gravity only acts in the vertical Direction gravity only affects the vertical motion so if we look at the Y position over time the ball is accelerating downwards due to gravity that acceleration causes the vertical velocity to increase the ball speeds up in the vertical Direction and Falls a greater distance during each second this should look familiar because it's the same as one-dimensional projectile motion sometimes called freeall this is a really important concept the vertical motion of the ball shot out of the cannon is identical to a ball falling straight down the initial veloc vity is horizontal so the initial Vertical Velocity is zero that's the same thing as being dropped from rest if we're only looking at the Y motion the balls have the same height and velocity at each time point and they hit the ground at the same time the fact that the ball is also moving to the right has no effect on its vertical motion so why is the horizontal motion different well the there's no gravity in the X Direction so there's no acceleration in the X Direction gravity pulls the ball down but it doesn't pull it to the right or left if the X acceleration is zero then the x velocity is constant it doesn't change over time so the ball moves the same distance in the X Direction during each second we could think of the ball's horizontal motion as being identical to a ball rolling on the ground at a constant velocity or if we turned gravity off the ball would move in a straight line at a constant speed the difference between this path and the ball's actual path is the amount the ball Falls due to gravity but the vertical acceleration has no effect on its motion in the X direction as an analogy here's a bird's eye view of a curling Stone sliding on ice with no friction which we used in some previous videos this is 2D motion on a horizontal plane so the X and Y directions are both horizontal on the ground if we give this Stone a push in the X Direction so we give it an initial x velocity it'll move in a straight line at a constant speed if there's no friction now imagine we attached a model rocket engine to the curling Stone when we light the engine it causes the curling Stone to accelerate if we point it in the negative y direction the Y velocity will increase and the stone will move further during each interval of time now what if we light the engine and at the same time we give the stone a push in the X Direction now the stone follows this curved path which is a combination of a constant velocity in the X Direction and acceleration in the y direction in this example the acceleration is caused by the rocket engine in projectile motion the acceleration is caused by gravity this picture will change a little bit when we have different initial values but this is how two-dimensional projectile motion works the ball has separate X and Y positions separate X and Y velocities and we use separate X and Y equations but we can take the X and Y velocities and combine them to get the actual velocity Vector of the ball at any time we can draw the components like this or we can draw them so they form a right triangle with the vector v is the instantaneous velocity Vector of the ball it shows the ball's actual velocity through the air and the direction of its motion at each instant this might be the first time we're seeing a velocity Vector on a curved path where the direction is changing notice that the velocity Vector does not trace the path of the object the path is curved but a vector is just a straight arrow the velocity Vector is always tangent to the path and it shows the direction of the Velocity at one instant so now that we have an idea of how things work let's use the equations and see what the motion graphs look like for this example here's the information we have first let's use this equation to find the Y position at each time Point like we mentioned in the last video the first variable in this equation is written as y final when we solve for the Y position at each time you you can think of each one as the final y position in the equation but we're just going to write them as y with no subscript at the point where T equals 0.5 seconds we can plug in 19.6 M for the initial y position 0 m/s for the initial y velocity 0.5 seconds for the time and 9.8 m/s squar for the Y acceleration that gives us 18.4 M for the Y position at 0.5 seconds we can do the same thing for each time Point by plugging in the same initial values and the different times at 1 second the Y position is 14.7 M at 1.5 seconds it's 8.6 M and at 2 seconds the Y position happens to be 0 m so we can see the Y position the height is decreasing as the ball Falls here's the Y displacement between each point the ball is moving further during each interval because it's accelerating here's the graph of the ball's y position versus time and the points we just found the horizontal axis is time in seconds and the vertical axis is the Y position in meters the line starts at the initial y position 19.6 M it decreases over time and it crosses the horizontal axis at 2 seconds that's when the Y position is zero and the ball hits the ground the graph is a curved line because the ball is accelerating in the y direction the graph and the actual path of the ball are parabolas but this graph is not showing us the two-dimensional path of the ball it's just showing us the Y position over time if the ball was falling straight down this graph would be the same it's just showing how the Y position is changing also this is a graph of this function which is the kinematic equation we used with the initial values and it's written as the Y position as a function of time now let's find the Y velocity at each time using this equation this tells us how the velocity is changing over time due to an acceleration when T equals 0 0.5 seconds we can plug in 0 m/s for the initial y velocity 9.8 m/s s for the Y acceleration and 0.5 seconds for time that gives us -4.9 m/s for the Y velocity at that time at 1 second the Y velocity is 9.8 m/s at 1.5 seconds it's -4.7 and at 2 seconds it's 9.6 here's what the Y velocity vectors look like as the ball Falls they're always negative because they point in the negative Direction but the magnitude is increasing the ball is speeding up if we look at the Y velocity at 1 second intervals it's changing by the same amount each second that amount is the acceleration 9 .8 m/s and here's the graph of the Y velocity versus time the horizontal axis is time but now the vertical axis is the Y velocity in m/s up is the positive direction and down is negative it starts at 0 m/s which is the initial y velocity the slope of the Velocity graph is the acceleration the change in velocity over the change in time which is 9.8 m/s/s here's the Y position y velocity and Y acceleration graphs together the Y acceleration is always a constant 9.8 m/s squared so it's a flat line the slope of the Velocity graph is the acceleration the acceleration is negative so the slope of the Velocity graph is negative and the slope is constant so the velocity is a straight line the slope of the position graph is the velocity the change in position over the change in time at 0 seconds the slope of the position graph is zero and the velocity is zero at 1 second the slope of the position graph is 9.8 which is the velocity at 1 second the velocity is changing so the slope slope of the position graph is changing which makes it a curve now let's look at the motion in the X direction there's no horizontal acceleration so we can use this equation to find the X position at each time at T = 0.5 seconds we plug in 0 m for the initial X position 10 m/s for the x velocity which is always the same as the initial x velocity and we plug in 0.5 seconds for time that gives us 5 m for the X position at 1 second the X position is 10 m at 1.5 seconds it's 15 M and at 2 seconds it's 20 M notice how the X displacement is the same for each interval because the x velocity is constant the ball moves 5 m to the right every half second here's the graph of the X position versus time now the vertical axis of the graph represents the X Direction the initial position is 0 m and as the ball moves to the right the line on the graph moves up this is how motion graphs are set up we always have time on the horizontal axis this is a graph of this function which is the kinematic equation we just used the Y inter is 0 which is the initial X position and the slope is 10 which is the x velocity notice how there's nothing special about this graph at 2 seconds when the ball hits the ground the line doesn't stop cross an Axis or reach some Maximum value if the ground wasn't there to stop the motion or if there was no gravity the graph would be the same and the ball would keep moving to the right forever so if we want to find the X position when the ball hits the ground which is the range of the projectile motion we first need to find the time it hits the ground using the Y Motion in this case that's at 2 seconds then we can find the X position at that time which is the range what about the Velocity in the X Direction well since there's no acceleration in the the X Direction the x velocity is always the same as the initial x velocity 10 m/s the magnitude or length of the x velocity Vector doesn't change over time here's a graph of the x velocity versus time it's not that interesting but it is another way to show that the x velocity is a constant 10 m/s and here's the graphs for the X position x velocity and X acceleration these are just the graphs for an object with a constant velocity so again they're not as interesting as the Y graphs the acceleration in the X direction is zero the velocity is always positive 10 m/s and the X position increases over time since the acceleration is zero the slope of the Velocity graph is zero and the velocity is constant so the slope of the position graph is constant and it's a straight line so that's what 2dimensional projectile motion looks like when an object starts at some height with a horizontal velocity now that we've looked at the motion let's answer a few questions how long is the ball in the air how far does the ball travel and what is the speed of the ball when it lands let's start with the first one we mentioned earlier that the time in the air is only determined by the Y motion so we're looking for the time when the ball hits the ground which is the time when the Y position is Zer the only equation we can use with the information we have is this one we can plug in 0 m for the final y position when the ball hits the ground 19 .6 M for the initial y position 0 m/s for the initial y velocity 9.8 m/ second squared for the acceleration and we want to solve for T we can simplify this equation add 4.9 t^2 to both sideside 4.9 take the square root of both sides and we get 2 seconds for T that's the time when the ball reaches 0 m and it's how long the ball in the air next how far does the ball travel this is asking for the displacement of the ball in the X Direction which is the range of the projectile motion how do we find that we know the ball is moving to the right at a constant velocity and now we know how long the ball is in the Air 2 seconds that time applies to the Y motion and the X motion that's how long the ball has to move to the right so we can use that time and the velocity to find the displacement so let's use the equation for the x velocity we can multiply both sides by time and then plug in 10 m/s for the x velocity and 2 seconds for the time this equation uses delta T but that's the same as T because the initial time is zero when we mult mply those together we get 20 M for the X displacement which is the range we'll talk more about range later in the video finally what is the speed of the ball when it lands are we looking for the x velocity or the Y velocity when a question asks for the speed or velocity of an object in 2D motion it's talking about the actual velocity Vector which is the combination or the Vector sum of the X and Y velocity components so the final velocity Vector is the combination of the final x velocity and the final y velocity speed is the magnitude of velocity which is the length of the Velocity Vector so if we know the components we can find the magnitude using the Pythagorean theorem we already know the final x velocity it's the same as the initial x velocity velocity 10 m/s so we need to find the final y velocity we could use this equation and plug in the time when the ball hits the ground or we could use this equation and plug in the final y position the first equation only works because we already found the time so let's use this equation we plug in 0 m/s for the initial y velocity 9 8 m/s squared for the Y acceleration 0 m for the final y position and 19.6 M for the initial y position if we simplify that and take the square root of both sides we get 19.6 m/s notice that this equation only deals with positive values for the velocities because they're squared in the equation so it's really giving us the final y speed or the absolute value of the final y velocity the Y velocity points in the negative Direction so the final y velocity would be 19.6 m/s so now we know the X and Y velocity components when the ball hits the ground remember from the lesson on vectors that we use the Pythagorean theorem to find the magnitude of a vector which is the hypotenuse so the final velocity is the sare root of the final x velocity squ plus the final y velocity squ we plug in 10 m/s and 9.6 m/s and we get 22 m/s for the magnitude of the final velocity which is the final speed all right so that's an example of two dimensional projectile motion where an object starts at some height with a horizontal velocity now let's look at one more example where an object has an initial velocity at an angle we'll go through this one faster a ball is launched from the ground with a velocity of 22 m/ second at an angle of 63° if the angle isn't described relative to the vertical or some other line we'll assume it's 63° above the horizontal or above the ground for this example we'll place the origin right at the starting point of the ball so the initial X and Y positions are both zero how is the ball being launched from the ground like we said before that doesn't really matter for projectile motion we just care about the initial height and the initial velocity we could say the ball was kicked into the air or maybe it was shot out of a cannon below ground level the point of a setup like this is that the initial and final Heights are both zero next what do we do with this velocity Vector in pretty much every case we need to find the components the initial x velocity and the initial y velocity remember a vector and its components form a right triangle so we can use the right triangle trig relationships the initial x velocity is the initial velocity time the cosine of this angle which is 10 m/s the initial y velocity is the initial velocity time the S of this angle which is 19.6 m/s cosine and S don't always go with X and Y just remember that cosine is for the component adjacent to the angle we're using and S is the component opposite from the angle so these are the initial X and y velocities when the ball is launched its actual velocity is 22 m/s in this direction but it's moving up at 19.6 m/s and it's moving to the right at 10 m/s at this initial time now we have all the initial values so what does this motion look like what's happening to the X and Y positions over time what about the X and Y velocities is there anything about this that's similar to the last example here's the ball's position at half second intervals the ball moves up reaches a maximum height and then falls down and it's moving to the right the whole time if we look at the horizontal motion the ball moves the same distance to the right during each interval just like before the x velocity is constant because there's no acceleration in this direction the ball is moving 10 m/s to the right the whole time so the ball moves the same amount each second it's the same motion as a ball rolling on the ground at 10 m/s it doesn't matter what's happening in the vertical Direction the horizontal motion will always be the same now let's look at the vertical motion During the period when the ball is moving up the distance between each height is getting shorter over time why is that gravity is acting downwards in the opposite direction as the velocity if the velocity is positive and the acceleration is negative the velocity will decrease over time and the ball slows down in the vertical Direction so the ball moves less vertical distance during each interval then the ball reaches a maximum height and falls back down as the ball Falls the Y velocity increases and the ball Falls a greater distance during each interval just like we saw in the first example so for the first half the ball moves up but it slows down in the y direction during the second half the ball falls down and moves faster in the y direction notice that the height of the ball is the same at equal times before and after the maximum height half a second before and after it's at this height 1 second before and after it's at this height since the initial and final Heights are the same this trajectory is symmetric the ball takes 2 seconds to move up and 2 seconds to fall down also notice that this is the same vertical motion as a ball that's launched upwards at 19.6 m/s just like with 1D projectile motion when the ball reaches the maximum height the Y velocity is zero because it's reversing Direction the ball is still moving to the right but it's not moving up or down at that moment so the horizontal velocity is constant but the Y velocity changes because of the acceler ation due to gravity if we combine the X and Y velocity components at each time point we can see the actual velocity Vector of the ball the X components stay the same but the Y component changes which changes the length and direction of the vector the vector always points in the direction of the ball's motion and when the ball is at the maximum height the Y velocity component is zero so the velocity is just the x velocity if we line up the vectors at each time we get a different view of how the velocity Vector is changing the x velocity stays the same but the Y velocity starts pointing up then gets shorter until it's zero then it flips and gets longer in the negative Direction so the vector changes length and it rotates now let's quickly run through the actual values of the positions and velocities then we can look at the motion graphs for this example let's start with the X motion we can use this equation to find the X position at each time Point like before we plug in 0 m for the initial X position 10 m per second for the x velocity and the times to get the x positions the x velocity is constant so the ball moves 5 m during each half-second interval here's the graph of the X position versus time it starts at 0 m and increases with a constant slope which is the velocity since the x velocity is constant it's always the same as the initial x velocity 10 m/s if we graph that it's just a flat line here's the position velocity and acceleration graphs for the X motion the position goes from 0 m to 40 m the velocity is always 10 m/s and the acceleration is zero these are the same graphs that we'd have for a ball rolling on the ground at 10 m/s since we're only looking at the motion in the X Direction now let's look at the Y motion we can use this equation to find the Y position at each time the initial y position is 0 m the initial y velocity is 19.6 m/s and the Y acceleration is 9.8 m/s squared when we plug in each time we get the Y positions notice how the Y position increases from 0 to 2 seconds then it decreases we don't need to calculate the up and down motions separately this equation covers the entire vertical motion the initial velocity is positive and the acceleration is negative and that's how the math works out here's the graph of the Y position versus time it starts and ends at 0 m and it reaches a maximum height of 19.6 M at 2 seconds this graph looks like the two-dimensional path of the ball but it's not it's just showing us the Y position over time this is the same graph as a ball that's launched directly upwards with the same initial y velocity we can find the Y velocity at each time using this equation the initial y velocity is 19.6 m per second and the Y acceleration is 9.8 m/s squ notice how the velocity starts at positive 19.6 it decreases to zero at 2 seconds then it becomes negative and increases in magnitude again this is how the equation works the initial velocity is positive and we subtract from it each second because the acceleration is negative here's the graph of the Y velocity versus time it starts at 19 .6 m/s and ends at 19.6 m/s it has a negative slope which is the acceleration and this is the same velocity graph as a ball launched directly upwards the velocity starts out positive it slows down to zero at the maximum height at 2 seconds then it becomes negative as the ball falls down and the velocity has the same magnitude at equal Heights except the direction is flipped here's all the graphs for the Y motion the slope of the position graph is the velocity so the position starts with a positive slope the slope decreases to zero then it increases in the negative Direction so how does this motion compare to the first example in that example the ball started at some height with a horizontal velocity that motion is just the second half of this motion it just starts 2 seconds later when the ball is at the maximum height the x velocity is 10 m/s and the Y velocity is zero they both take 2 seconds to fall from the maximum height to the ground here's the X and Y positions and velocities for the second example if we start the clock 2 seconds later we get the values for the first example if we start with the Y motion graphs from the second example and we shift them 2 seconds to the left we get the graphs from the first example so the ball starts at a height of 19.6 M and A Y velocity of zero and the same is true for the X motion graphs the X velocities are the same but in the first example the ball is in the air for half the time so it only travels half as far before hitting the ground now let's wrap up this example by finding the speed of the ball when it lands just like the first example the final velocity Vector is the combination of the final x velocity and the final y velocity we know the final x velocity so let's find the final y velocity using this equation we plug in 19.6 m/s for the initial y velocity 9.8 m/s squar for the Y acceleration 0 m for the initial y position and 0 m for the final y position when it hits the ground since the initial and final y positions are the same the terms on the right are multiplied by 0 the final y velocity squared is equal to 19.6 S so the final y velocity is 19.6 m/s again this equation actually gives us the speed because V is squared the final y velocity is 19.6 m/s since it points downwards now that we have the final X and Y velocities we can use the Pythagorean theorem to find the speed which is the magnitude of the final velocity Vector when we do that we get 22 m/s for the final speed of the ball this is the same as the initial speed when the ball is launched that's because the initial and final Heights are the same and the trajectory is symmetric so we could have just answered this question by remembering that concept finally what if we wanted to answer this question how far does the ball travel let's talk about one of the important Concepts in projectile motion which is range range is how far the object travels in the horizontal Direction it's the displacement in the X Direction so what determines the range how do we find Delta X here's the equation for the x velocity if we know VX and delta T we can find Delta X VX is the horizontal velocity which is always the same as the initial horizontal velocity delta T is the amount of time the object is in the air which only depends on the Y motion we actually did this in the first example but let's outline the steps for finding the range first we need to find the initial X and Y velocities the components of the initial velocity Vector for this examp example the initial x velocity is 10 m/s and the initial y velocity is 19.6 m/s second we need to find the time in the air using the Y motion if the initial and final Heights are different like if the object is launched off a cliff we need to use this equation to find the time but if the initial and final Heights are the same like in this example the trajectory is symmetric so we know the final y velocity then we can use this equation for acceleration the change in the Y velocity is the final y velocity minus the initial y velocity we can rearrange the equation so time is on the left then plug in the values for the velocities and the acceleration and we get 4 seconds that's the duration of the motion the amount of time the ball is in the air third we use that time to find the range Delta X the x velocity is the initial x velocity 10 m/s and the time is 4 seconds that gives us 40 M for the range here's all three steps it's always good to understand how to solve a problem like this but we could combine these steps and come up with an equation for the range using any velocity in angle we'll move through this quickly so feel free to pause at any point also see if you can derive this equation on your own based on these steps we'll use this equation to find the time in the air we replace the initial y velocity with the initial velocity times the S of theta and replace the acceleration with negative G then we rearrange things so we can use the quadratic formula we can keep y initial and Y final or we can say y final is zero we plug these coefficients into the quadratic formula simplify that and we get an equation for T the time in the air the range is the x velocity times the period of time we replace the x velocity with the initial velocity time the cosine of theta and we replace t with the equation we just found so this is the range equation for any projectile motion but if the initial and final Heights are both zero then we can simplify this a lot more y initial is zero so the square root term simplifies to VI * the S of theta then we can use this trig identity 2 * cosine * sin equal the S of 2 Theta then we end up with this this is the range equation you might see more often keep in mind that this only applies if the initial and final Heights are equal like if the object starts and ends at the ground one more question that shows up a lot is this what launch angle results in the longest range so what value of theta will give us the maximum value of delta X here's the graph of of this equation using 22 m/s for the initial velocity the angle Theta is on the horizontal axis from 0 to 90° and the vertical axis is the range Delta x no matter what the initial velocity is an object launched from the ground will travel the farthest when it's launched at 45° if the angle is lower the x velocity is faster but the object spends less time in the air if the angle is higher the object spends more time in the air but the x velocity is slower another important thing to know is that the range is the same at equal angles above and below 45° so the ball travels the same distance if it's launched at 30° or 60° which are both 15° away from 45 and the ball travels the same distance at 15° and 75° but what if the object is launched from an initial height then the initial and final Heights are not the same so we use the general range equation and the graph looks different it's no longer a regular sign curve more importantly if the initial height is greater the maximum value shifts to the left if we keep the initial velocity of 22 m/s but the object starts at a height of 20 M the angle that gives us the maximum range is 36.6 Which is less than 45 so that's a closer look at range in projectile motion we covered two examples in this video but there's other scenarios for 2D projectile motion the only difference is the initial and final height and the angle of the initial velocity but they all follow the same rules that we covered in this video and we would solve problems the same way in each case to wrap up 2D projectile motion here's what the motion graphs would look like for each scenario this is the second example that we did where an object starts and ends at the ground we'll say the origin is on the ground at the initial X position the X motion graphs are the same for every scario so we'll just show them for this example the acceleration is zero the velocity is constant and the X position increases linearly and here's the Y motion the Y position starts and ends at zero and the Y velocity goes from positive to negative the object reaches the maximum height and zero y velocity halfway through the motion the Y acceleration graph will be the same in every case 9.8 m/s squ if up is positive if the final height is greater than the initial height the Y motion looks like this they're basically the same but they stop earlier the trajectory is not symmetric so the time when it reaches the maximum height and zero velocity is not in the middle it's later if the initial height is greater than the final height and the launch angle is upwards the graphs look like the last example but they're flipped the Y position starts at some height and ends at zero it reaches the maximum height and zero y velocity in less than half of the total time if the launch angle is zero so the initial velocity is horizontal the Y position starts at the maximum height with zero slope and then it decreases to zero and the Y velocity starts at zero and increases in the negative Direction and finally if the launch angle is below the horizontal the Y velocity starts out negative and increases in the negative Direction the Y position starts with a negative slope and goes to zero all right so that's everything we're going to cover in this video let's do a summary of two-dimensional projectile motion in 1D projectile motion an object only moves up and down in the y direction in 2D projectile motion an object moves in the X and Y directions we use the same y equations but we don't use x equations with acceleration the vertical acceleration is 9.8 m per second squared if up is positive and the horizontal acceleration is zero so the horizontal velocity is constant there's a few different scenarios for 2D projectile motion but they all work the same way the horizontal motion and the vertical motion the X and Y motion are completely independent the object has separate X and Y positions and velocities which use separate equations when working with projectile motion we're usually given the initial height and the initial veloc Vector we need to find the components of the vector which are the initial X and Y velocities then we can study the horizontal and vertical motion separately there's no horizontal acceleration so the x velocity is constant it's always the same as the initial x velocity and the object moves the same horizontal distance every second the vertical motion is affected by gravity so the acceleration in The Wider direction is always 99.8 m/s s downwards in 2D projectile motion the Y motion is the same as 1D projectile motion the Y velocity slows down as the object moves up the Y velocity is zero at the maximum height and then the Y velocity speeds up as the object falls the object will be at the same height at equal times before and after the maximum height and the Y velocity is the same at the same height but the direction is reversed the actual velocity Vector of the object at any time is the sum of the X and Y velocity components at that time at the maximum height the Y velocity is zero so the velocity Vector is just the x velocity the motion graphs for the X direction are always the same the X acceleration is zero the x velocity is constant and the X position increases linearly and we can use this equation to find the X position at any time the motion graphs for the y direction will depend on the initial and final points but in each case the acceleration graph is a constant 9.8 m/s squ if up is positive the velocity graph is a straight line and the slope is the acceleration the velocity curve crosses the horizontal axis when the Y velocity is zero and the object is at the maximum height and we can use this equation to find the velocity at any time or the time when the velocity is zero the Y position graph is curved because there's acceleration the slope at a point on the position graph is the velocity at that time and we can use this equation to find the Y position at any time the range of the projectile motion is Delta X how far the object travels in the horizontal direction to find the range we need the initial X and Y velocities then we find the amount of time the object is in the air using one of these equations then we plug that time into the x velocity equation to find the range if the initial and final Heights are the same like if the object starts and ends at the ground then we can use this equation to find the range a launch angle of 45° results in the maximum range and the range is the same for two angles that are the same amount above and below 45° and if the object starts at a greater height the optimal angle is less than 45° all right that's it for this video thanks for watching and I'll see you in the next one