hey everyone in this video we're going to cover some Concepts and tips for solving projectile motion problems and then we'll walk through some examples we already solved some problems in the lesson videos so these will be a little more difficult and you should watch the lesson videos first so let's start with some Concepts that are used in projectile motion problems if you need to review anything you can watch the lesson videos and check out the previous lessons here's some of the important things to remember about projectile motion X and Y are independent so an object's Motion in the X direction is independent from its motion in the y direction the acceleration due to gravity G is 9.8 m/s squ downwards so the vertical acceleration a y is either positive or negative G depending on which direction we choose to be positive the horizontal acceleration is zero so the horizontal velocity VX is constant and it's always the same as the initial horizontal velocity and if an object moves up and then falls down the vertical velocity VY is zero at the maximum height we also need to understand the graphs for projectile motion next we need to know how to use the kinematic equations for the X and Y directions since there's no acceleration in the X direction we only use the first equation that relates the X position velocity and time since we do have acceleration in the y direction we won't use this first equation unless we want to find the average velocity between two points and remember that Delta means final minus initial which applies to any variable you might see some equations use T for time and others have delta T in most cases we just say the initial time is 0 seconds so T and delta T mean the same thing and for two-dimensional projectile motion we need to know how to work with vectors and trigonometry a vector and its two components form a right triangle so we can use right triangle trig to find the components the magnitude or the angles it's better to know how to apply the trig relationships instead of memorizing if x or y goes with s or cosine because that depends on the angle that we're using now let's cover a few tips for solving these types of problems tip number one choose the origin and the positive X and Y directions we always want to establish the origin before starting a problem that's the point where X and Y equal Z and which directions are positive it's common to set up the origin on the ground and so the initial X position is zero once you set up the origin stay consistent with the positive and negative directions and the positive and negative values in the equations the vertical acceleration is G the acceleration due to gravity which is 9.8 m/s squ downwards so if we say up is the positive y direction gravity acts in the negative Direction and a y equals g and if down is positive gravity acts in the positive direction so a y equal positive G tip number two the X motion and the Y motion are independent this only applies to two dimensional motion but an object's Motion in the X Direction and the y direction are independent from each other if we're given the initial velocity and an angle we usually need to find the X and Y components of the initial velocity Vector which will be the initial x velocity and the initial y velocity remember that cosine is for the component that's adjacent to the angle and sign is for the component opposite the angle once we have the initial X and Y velocities we can use the 1D kinematic equations for each Direction separately but remember that time is the same for both directions so the value of T is the same in the X and and Y equations at the same moment in time and we can use the time variable as a link between the X and Y equations and when working with 2D projectile motion keep the variables and equations for the X and Y motions organized there's a lot of variables in subscripts so we want to be clear about which values we're given and what we're trying to find and tip number three some problems require multiple steps this appes to any physics topic but when starting a problem write out all the known and unknown values for the X and Y directions then we can look through the list of equations that apply and ask ourselves which kinematic equation has the variables we already know and the variable we're looking for if we can't solve a problem using one equation then we should start with the variable we're looking for and work backwards and use multiple equations to find intermediate variables for example let's say we already wrote down all the variables we know and the question wants us to find the final X position if we look at the equations the first one is the only equation that has that variable let's say we know the initial X position and the time but we don't know the x velocity so we need to find that first maybe we can use a trig relationship because we know the initial angle Theta but if we don't know the initial y velocity then we need to find that maybe using another kinematic equation if we know all of the other variables then we can start with that equation and work backwards we find the initial y velocity then we find the initial x velocity and then we can find the final Exposition in general it helps to write everything out and think through the problem before starting so those are some Concepts and tips for solving projec St motion problems now let's walk through some examples a ball is thrown directly upwards at 5 m/s the ball is released 1.5 m above the ground how fast is the ball moving when it lands on the ground as always let's read the problem a second time pick out the important information and draw a picture for projectile motion some important information could be the initial and final positions velocities or times a ball is thrown directly upwards at 5 m/s the ball is released 1.5 m above the ground how fast is the ball moving when it lands on the ground let's draw a picture the ball starts at some height above the ground with an upwards velocity so it moves up some distance and then falls down and hits the ground for most problems we'll say that up is the positive y direction and Y equals 0 at the ground we'll say the initial point is when the ball starts at this height and the final point is when it hits the ground so the initial y position is positive 1.5 M and the final y position is 0 m the initial velocity is POS 5 m/s which is also the initial y velocity because it points point in the y direction the question says how fast is the ball moving when it hits the ground which means we're looking for the final vertical speed which is the magnitude or absolute value of the Velocity since this is projectile motion the acceleration due to gravity is G 9.8 m/s squ which acts downwards since we chose up to be positive then the vertical acceleration a y would be G 9.8 m/s s and since the ball only moves vertically this is 1D projectile motion and we only use the variables and equations for the y direction also we mentioned this in the lesson videos but projectile motion really describes the motion of the object's Center of mass which is just a point so we could just draw a DOT to represent the ball if you do draw the object ignore the fact that it has some width and height and just think about the position of its Center so now we have a picture and we know which values were given and what we're trying to find you can include the values in the picture or just write them somewhere else like we did on the left so next what equations can we use to find the final speed of the ball let's look at the kinematic equations in the y direction remember since there's vertical acceleration we don't really use the equation for average velocity the last equation has the variable we're looking for and we have values for everything else so we can use that equation we could use the third equation but we would need to know the time it hits the ground and we could use the fourth equation to find that time because we know all the other variables in some Physics problems there's more than one way to solve it so let's try both first let's use this equation to find the final speed of the ball we plug in 5 m/s for the initial y velocity 9.8 m/s squared for the acceleration 0 m for the final y position and 1.5 M for the initial y position we can take the square root of both sides plug that into our calculator and we get 7.38 m/s for the final speed of the ball when y equal 0 m you can think of this equation as using speed instead of velocity because V is squared in the equation so even if the velocity is negative it'll be positive when you square it so this answer is the final speed of the ball which is what we want the final velocity would be 7.38 m/s because the ball would be moving in the negative y direction now let's solve this using the other equation we know the initial velocity and the acceleration but we need to know the time the ball hits the ground which we can find using this equation the final y position is 0 m the initial position is 1.5 M the initial velocity is 5 m/s and the acceleration is 9.8 m/s squar we can either plug this into a calculator that can automatically solve for t or we need to use the quadratic formula in that case we need to simplify and rearrange this equation so it's in this form you can move things around so the terms are in the same order but the important part is that one side of the equation is zero and we have three coefficients or numbers a b and c a is the number multiplied by t^2 which is -4.9 B is the number multiplied by T which is 5 and C is the number all by itself which is 1.5 the terms have to be added together which is why a is 4.9 then we can plug a b and c into the quadratic formula when we calculate that using the plus sign and the minus sign we get two values for T 0.24 seconds and positive 1.26 seconds we covered this in the lesson videos but we're going to ignore negative time values so the time when the ball hits the ground is positive 1.26 seconds now we can plug this value into the other equation to find the final velocity the initial velocity is 5 m/ Second the acceleration is 9.8 m/s squared and the time is 1.26 seconds that gives us - 7.35 m/s for the final velocity now let's check that our answer makes sense first of all did we get the same answer both times actually the first answer we got was 7.38 m/s so why are they different well we found the first answer by plugging in exact values that we were given for the second answer we calculated the time ourselves and we rounded that to two decimal places and then plugged that into another equation if we rounded time to three decimal places instead we'd get 7.38 m/s so if you're rounding a number in the middle of a problem make sure you keep enough decimal places so your final answer is accurate or if possible keep everything as variables and only plug in numbers at the end now let's check our answer the question is asking for the final speed which is the absolute value of velocity and that's what we got the next thing to check is the units our answer is in m/s which is the unit for speed the problem only used meters and seconds it didn't use other units like kilometers or hours and we don't have to convert anything so the units check out finally does the magnitude of the answer make sense we got the same value using two different methods so it's probably right the ball started with a speed of 5 m/s upwards so when it hits the ground it's definitely moving and the final speed isn't zero should it be greater or less than 5 m/s based on what we learned in the lesson videos the ball would have the same speed at the same height on the way up and down so when it's at 1.5 M on the way down its speed would be 5 m/s downwards then it falls further and keeps accelerating so its speed at the ground should be greater G than 5 m/s which is what we got once we cover more topics we could also try solving this using other Concepts like the conservation of energy and make sure we get the same answer but for now let's call this one solved and take a look at the next problem a coin is flipped into the air so it moves straight up and down the coin is caught 1 second later at the same height that it was flipped from what was the initial speed of the coin when it was flipped all right so let's read that again and pick out the important information a coin is flipped into the air so it moves straight up and down the coin is caught 1 second later at the same height that it was flipped from what was the initial speed of the coin when it was flipped so let's draw a DOT for the coin and we can draw some arrows to show that it moves up and then falls down to the same height again the coin only moves in the y direction so this is onedimensional projectile motion let's set up the Y AIS so y equals z at the height the coin is flipped and say up is positive that means the initial and final y positions are both 0 m we know the amount of time between those points is 1 second and since up is positive the vertical acceleration is netive G and we're trying to find the coin's initial vertical speed so which equation can we use to do that the second equation would work we know the initial and final positions the time and the acceleration so we could find the initial Vertical Velocity we could also use the first equation if we apply some of the stuff we learned about projectile motion the velocity of the coin at the maximum height is zero and since it takes 1 second for the coin to move up and back down to the same height we know the time to reach the maximum height would be half of that so if we treat the maximum height as the final point we could use this equation but let's start by solving this with the other equation where the final point is when the coin falls back down to the same height we plug in 0 m for the final position 0 m for the initial position 1 second for time and 9.8 m/s squ for the acceleration if we simplify that and rearrange things we get 4.9 m/s for the initial Vertical Velocity the speed would be the absolute value of that but it's already positive so this is also the initial vertical speed now let's solve this using the other equation this time we're saying the final point is when the coin reaches the maximum height the coin takes 1 second to move up and fall down to the same height since the motion is symmetric the time at the maximum height is half of the total time so 0.5 seconds the final speed at the maximum height is 0 m/s the acceleration is 9.8 m/s squared and the time is 0.5 seconds when we simplify that we get the same answer 4 9 m/s now let's check our answer the unit is correct and the speed should be positive how can we double check the value we already solved this using two different equations so it's probably right but another thing we can do is plug the answer back into the equations we used along with the other values were given and make sure the equation checks out if we do that the right side simplifies to 0 m so the equation is true both sides are equal which means the answer is right now let's check out the next problem a cannonball is launched upwards from an initial height of 0.5 M and its motion is shown in the graph what maximum height does it reach okay so this problem involves a motion graph let's figure out what we're looking at the horizontal axis is time in seconds and the vertical axis is velocity in m/s so we have a graph of the ball's velocity over time and the graph is giving us the coordinates or values for two specific points let's read the problem again and find the important information a cannon ball is launched upwards from an initial height of 0.5 M and its motion is shown in the graph what maximum height does it reach so the ball is only moving in the y direction we know the initial position and we want to find the Y position at the maximum height now let's draw a picture the ball's initial height is 0.5 M so we'll say y equals 0 at the ground and up is positive the ball moves upwards and reaches a maximum height we also know that whenever an object is at the maximum height the vertical velocity is Z and we know the vertical acceleration will be G now what information can we get from the graph at the initial point when T equals 0 the velocity is 25 m/s which is VY initial then the velocity is 0 m/s when t equal 2.55 seconds that's the point when the ball is at the maximum height the ball starts moving upwards with a positive velocity but it's slowing down it reaches 0 m/s at the maximum height and then the velocity becomes negative and speeds up as the ball falls down so which equations can we use to solve this we could use this equation and say the final point is when the ball is at the maximum height so we would solve for y final and we know everything else or we could use this equation which also includes y final and variables that we already know let's try using this equation first the final point is at the maximum height so we plug in 0.5 M for the initial position 25 m/s for the initial velocity 2.55 seconds for time and 9.8 m/s squar for acceleration that gives us 3239 M for the final y position which is the maximum height now let's solve this using the other equation instead of plugging in numbers at the beginning let's try rearranging the variables first to isolate the final position and then plug in numbers at the end first we can subtract VY initial squared IDE 2 a y then add y initial to both sides so y final is by itself now we can plug in the values that we have for each variable when we do that we get 32.3 n m for y final which is the maximum height that's the same answer we got before in the first equation we used the ball's initial height initial velocity the acceleration and the time it took to reach the maximum height but you might have noticed that with the second equation we didn't need to use the time at all so we could have solved this problem even if we weren't given the second point on the graph how are we able to find the maximum height only given the initial height and velocity well in the second equation we also used the fact that the velocity is zero at the maximum height the acceleration is 9.8 m/s squared which is the slope of the Velocity graph so if the initial velocity is fixed and the slope is fixed the velocity must be zero and across the horizontal axis at a fixed time so one equation uses the final time and one equation uses the final velocity but either way this is a reminder that a problem might give us more information than we need so now let's check our answer we got the same value using two equations and we have the correct unit for position or height so this is probably right now let's check out the next problem a ball is thrown horizontally out of an 8 m High window with a speed of 3 m/s at what angle does the ball impact the ground what is the angle of its velocity Vector relative to the ground all right now it sounds like we're dealing with two-dimensional projectile motion let's read that again and figure out what information were given a ball is thrown horizontally out of an 8 m High window with a speed of 3 m/ second at what angle does the ball impact the ground what is the angle of its velocity Vector relative to the ground so let's draw a picture and figure out what's happening first let's draw the ground and the ball starting at a height of 8 m the ball is thrown horizontally so we'll draw the initial veloc vity Vector pointing sideways the ball is going to move to the right and fall down so we can draw the ball when it hits the ground and the path of the ball we're trying to find the angle of the Velocity Vector when the ball hits the ground so here's the final velocity vector and we can label the angle of the vector relative to the ground as Theta final since the ball is moving horizontally we're dealing with two dimensional projectile motion so we need to set up the X and Y AIS we usually place the origin on the ground at the initial horizontal position right is the positive X Direction and up is the positive y direction you can draw the origin with the X and Y axis together or if the picture gets too crowded you can draw each axis separately now we can write out all of the given information keeping the X and Y variables separate first we're told the initial velocity is horizontal and has a magnitude of 3 m/s which is the speed if we want we could write that the angle of the initial velocity is zero because it's horizontal based on the axes we set up the initial X position would be 0 m we don't know the final X position we know the initial x velocity is 3 m/s because the initial velocity points in the X direction we also know the X acceleration is zero for projectile motion so the x velocity is constant and the final x velocity is also 3 m/s the initial y position would be 8 m the final y position at the ground would be 0 m the initial y velocity is 0 m/s because the initial velocity Vector does not have a y component we don't know the final y velocity but we know the Y acceleration is G 9.8 m/s squ now that we have everything set up which equations can we use to solve this let's think about how we can find the angle of the final velocity Vector we know that in two-dimensional motion the velocity Vector has X and Y components so if we draw the final velocity vector Vector the horizontal component is the final x velocity and the vertical component is the final y velocity we already know the final x velocity but we don't know anything else about this Vector so how can we find the angle since this is a right triangle we can use some right triangle trig if we knew the final y velocity we could use the inverse tangent function to find the angle or if we knew the magnitude of the final velocity which is the hypotenuse we could use the inverse cosine function but we would need to find the Y component anyway in order to find the hypotenuse so let's find the final y velocity and then use the inverse tangent function to find the angle so which equation can we use to find the final y velocity this equation has that variable and we know everything else so let's use this we plug in 0 m/s for the initial y velocity or the speed NE 9.8 m/s squared for the acceleration 0 m for the final y position at the ground and 8 m for the initial y position we can simplify that a little and take the square root of both sides and we get 12.52 m/s now we have have a right triangle where the X component is 3 m/s and the Y component is 12.52 m/s technically the Y component is negative but we only plug positive values into the inverse tangent function because we're just dealing with geometry and triangles have positive side lengths so we can plug in 12.52 for the opposite side three for the adjacent side and we get 76.5 de also make sure your calculator is set to degrees instead of radians so that's what the question is asking for the angle between the final velocity vector and the ground this opposite angle is also 76.5 de if you want to think about that as the angle the ball impacts the ground now let's check that our answer makes sense it would have to be a value between 0 and 90° because it's this angle in inside the right triangle 0° means the final velocity is horizontal and 90° means the final velocity is vertical so this is somewhere in between if we want we could use more decimals and plug this value into the equations we used to see if everything checks out but we'll move on to the next problem a golf ball is hit into the air with a speed of 30 m/ second and an angle of 60° there is a 10 m tall fence 70 M away from where the ball was hit does the ball make it over the fence all right what information are we given a golf ball is hit into the air with a speed of 30 m/s and an angle of 60° there is a 10 m tall fence 70 M away from where the ball was hit we're going to assume the ball is hit from the ground so let's draw the ground the ball the initial velocity vector and the fence the fence is 70 M away from the starting point and it's 10 m tall let's place the origin right at the initial position of the ball so the ball starts at xal 0 and y equal 0 and we know the vertical acceleration is G now let's write out the values that were given the initial velocity is 30 m/s at an angle of 60° since we're not given any other information we'll assume that angle is relative to the ground X initial is 0 m and Y initial is 0 m based on the origin we chose what would be the final X and Y positions are they 70 M and 10 m we'll come back to that in a second we're not given the initial X and Y velocities but we're given the magnitude and angle of the initial velocity vector and the initial X and Y velocities are the components when we're given the initial velocity Vector we almost always start by finding the components 60° would be the angle between the vector and the X component which is parallel to the ground using this angle the initial x velocity would be 30 m/s time the cosine of 60° which is 15 m/s and the initial y velocity would be 30 m/s * the S of 60° which is 25.98 m/s but let's leave them as VI * the cosine and S of 60° so we can plug in the exact numbers at the end instead of rounding them now we know the x velocity is constant so the final x velocity at any point is just the initial x velocity and we don't know the final y velocity so the question is asking whether the ball makes it over the fence we don't know the exact path of the ball it could go over the fence or it could fall short in terms of the physics variables we're using what does it mean for the ball to make it over the fence one way we could solve this is by finding the height of the ball when it has the same hor horizontal position as the fence if the ball is higher than 10 m at that point then it makes it over the fence so in that case we're looking for the final y position or height when the final X position is 70 M the same as the fence the other way we could solve this is by finding the horizontal position of the ball when it's at a height of 10 m if the ball is to the right of the fence at that height it made it over and if it's to the left at that height it falls short so in that case we're looking for the final X position when the final y position is 10 m we'll try both methods but for these types of problems the first method is probably easier so which equations can we use to find the final y position when the final X position is 70 M the first y equation doesn't include position the second equation does and we know everything else except the time so maybe we can find that the third equation includes position but we don't know the final y velocity so let's use the second equation how would we find the final time if we look at the X equation we know the initial and final x positions and the x velocity so we can use that to find the time then we can plug that into this y equation to find the final y position and if we think about the motion these steps make sense we're choosing an X position and we want to find the Y position at that point there's no equation that directly relates the X and Y positions at least not in this course so we're taking that X position 70 M then finding the time when the ball is at that position and then we're finding the position at that time so let's start by finding the time remember this equation is just the velocity equation rearranged so we can use either one if we take the velocity equation and multiply both sides by delta T and then divide both sides by VX we have delta T on the left then we can plug in 70 M for Delta X or 70 M - 0 m if you want to use the final and initial positions then we plug in the value we found for VX which is the initial x velocity 30 m/s time the cosine of 60° that gives us 4667 seconds for the time it takes the ball to travel 70 M now we can find the final y position at that time using this equation we plug in 0 m for the initial y position 30 m/s time the S of 60° for the initial y velocity 4.66 7 seconds for time and 9.8 m/s squar for acceleration we rounded time to three decimal places so that our final answer is more accurate but we also could have plugged in that full expression for time instead of calculating it but the equation would get pretty long so when we calculate this we get 18.5 M for the final y position that's the height of the ball at 4.66 7 seconds and it's the height of the ball at an X position of 70 M which is the position of the fence the fence is 10 m tall and the ball is higher than that so the answer would be yes the ball does make it over the fence now let's solve this problem the other way by finding the final X position when the ball is at a final y position of 10 m now the final time is determined by the Y position so we start with this equation to find the time we plug in 10 m for the final y position 0 m for the initial y position 30 m/s time the S of 60° for the initial y velocity and 9.8 m/s squ for the acceleration to solve for T we need a calculator with a solve feature or we need to use the quadratic formula but here's what we get for T 0.418 seconds and 4.88 4 seconds remember that the graph of this function is a parabola and there's always two times when an object is at a certain height unless it's the maximum height if we look at the path of the ball it's at a height of 10 m as it moves up and when it falls down so let's find the x positions at both times using this equation we plug in 0 m for the initial X position 30 m/s time the cosine of 60° for the x velocity and both values for time that gives us 6.3 M and 73.3 M those are the two horizontal positions where the the ball is at a height of 10 m the fence is at a horizontal position of 70 M which is between those two points so looking at the picture if the ball is at a height of 10 m to the left and right of the fence it must be higher in the middle so it makes it over again solving it the other way might be easier to think about but either way works now let's check our answer we got the same result solving it two different ways so it's probably right we could plug the final x positions and the final y position from the other method back into the equations to check if the equations are true another thing we can do if we have enough time and we really want to be sure we got it right is to walk through the problem a second time to see if we get the same values again and check that we didn't make any mistakes with the math now now let's move on to the last problem a ball is kicked into the air at an angle of 50° if the ball is in the air for 3 seconds how far does it travel so that's not a lot of information let's read that again and think about how to solve this a ball is kicked into the air at an angle of 50 de if the ball is in the air for 3 seconds how far does it travel so let's assume the ball starts and ends at the ground we'll draw the ground the ball the initial velocity vector and the path through the air we'll place the origin where the ball starts let's also draw the initial velocity Vector over here with the angle and the two components the initial x velocity and the initial y velocity so the ball has some initial velocity we know how long it's in the air and we're trying to find how far it travels which is the range of the projectile motion we can call that Delta X or since the initial X position is zero Delta X is the same as X final we don't know the magnitude of the initial velocity but we'll write down that the angle is 50° the amount of time in the air is 3 seconds and the initial X and Y positions are 0 m we don't know the final X position that's what we're trying to find but since we're assuming the ball ends at the ground the final y position is 0 m we don't know the X and Y velocities but we know the X acceleration is zero and the Y acceleration is G so we're not given a lot of information how can we find the range of the ball and what equations can we use well the only equation that includes the X position is this one we know the initial X position and we know the time but we need to know the x velocity so how can we find that we're given the angle of the initial velocity Vector so if we could find the magnitude or the Y component then we could use some trig to find the X component we need to know the Y component in order to find the magnitude so let's just find the initial y velocity and use the tangent relationship to find the x velocity then we can use that to find the range so how do we find the initial y velocity we're given the amount of time the ball is in the air we learned in the lesson videos that the time in the air is determined by the Y motion it's just like onedimensional projectile motion so this ball has some initial y velocity it moves up and then it falls down if we know the initial and final y positions and the acceleration then there's only one initial y velocity that would cause the ball to be in the air for 3 seconds if it had a faster or slower initial Vertical Velocity it would be in the air for more or less than 3 seconds there's more than one way to solve this but if we look at this equation we know the initial and final y positions we know the time and we know the acceleration so we can solve for for the initial y velocity we plug in the values that we know and the only unknown is the initial y velocity if we simplify and rearrange that equation we get 14.7 m/s for the initial y velocity now we can use that to find the x velocity tan of theta equals the opposite side / the adj side So Tan of 50° equals the initial y velocity over the initial x velocity if we rearrange that we can plug in 14.7 m/s for the initial y velocity and we get 12.33 m/s for the initial x velocity now we can take that and plug it into this equation along with the initial X position and the time and we get 37 M for the final X position again that's the same as Delta X because the initial X position is zero so only using the initial angle and the time in the air we found the range which is 37 M how can we double check our answer based on the given information we can't really think through whether this value makes sense but like we mentioned before we can plug this this answer into the equations we used and work backwards to see if we get one of the given values so let's start with the range and see if we get 3 seconds for time using this equation we plug in 37 M for the final X position then using the angle of 50° and the tangent relationship then if we plug that into this equation and solve for time we get 3 seconds which is the value that we were given so our final answer should be right these values worked out well but if you do this keep in mind that you're plugging in rounded values so you might be off a few decimal places but you should be able to tell that it's close enough or you can use more decimal places and check it again so that's an example of a problem where we're not given a lot of information and we need to find different values using multiple steps all right that's it for this video thanks for watching and I'll see you in the next one