Understanding Projectile Motion Fundamentals

This video is going to be an updated version to an earlier video that I created on projectile motion. But let's go over some basics. The first thing you need to know are the kinematic equations. At constant speed, d is equal to vt. Displacement is equal to velocity multiplied by the time. Now under constant acceleration, displacement is equal to the average velocity multiplied by time. And the average velocity is basically the average of the initial and the final velocity. You add up the initial and the final and you divide it by 2 if the acceleration is constant. Now you also have some other equations. V final is equal to V initial plus AT. So the final velocity is equal to the initial velocity plus the acceleration multiplied by the time. Also, the square of the final velocity is equal to the square of the initial velocity plus 2 times the product of the acceleration and the displacement. And there's another equation. Displacement is equal to V initial t plus 1 half a t squared. Now, the displacement is the change in position. It can be the final position minus the initial position along the x-axis, or it can be along the y-axis. So if we replace D with Y final minus Y initial, we can get another equation that perhaps you've seen in your physics course by now, which looks like this. The final position is equal to the initial position plus V initial T plus 1 half A T squared. Now, D can represent displacement, or sometimes you can use it to find distance. Distance and displacement are the same if the object moves in one direction and doesn't change direction. Whenever the object changes direction, displacement and distance are different, so you've got to be careful in order to solve it. But if it's moving in one direction, distance and displacement, they're the same in that case. Now, let's say if we have a ball, and we're going to kick it off the ground, it goes up, and then it's going to go down. Gravitational acceleration is negative 9.8 meters per second squared. For this example, we're going to round it, and we're going to use negative 10, just to keep things simple. So, let's say at t equals 0, I'm going to put the time inside the ball. Let's say the horizontal velocity, Vx, we're going to say it's 7, and the vertical velocity, Vy, we're going to say it's 30. Now, one second later, what do you think the horizontal and the vertical velocity will be? What about two seconds later, and then three seconds later, and so forth? So one second later, I should probably put that in a different color, the horizontal velocity will not change. It's going to remain 7. The vertical velocity will change. And how much will the vertical velocity change by? The vertical velocity changes by the gravitational acceleration. Acceleration tells you how much the velocity changes every second. Since the gravitational acceleration, which is the acceleration in the y direction, it changes about 10. Every second the Vy is going to decrease by 10 every second so one second later. It's going to be 20 2 seconds later the Vx is going to be the same, but Vy is going to be 10 3 seconds later Vy is now 0 at the top. It's not going up anymore. You only have horizontal motion, but Vx is still 7 Then Vy becomes negative But Vx will always remain 7. Because Vx is constant, the acceleration in the x direction is 0. A projectile, by definition, is only under the influence of gravity. So if you throw a pen, once you release it from your hand, it's a projectile. A flying bird is not a projectile. It's in the air, but it has the force generated by its wings acting on it. So it's not a projectile. Projectiles can only have gravity acting on it. Air resistance is ignored. So this is going to be 2, 3, 4 seconds later, Vy is going to be negative 10. 5 seconds later, negative 20. 6 seconds later, negative 30. Now, here's a question for you. What is the vertical velocity and the speed 5 seconds later? The vertical velocity, Vy, is negative 20. The speed is positive 20. Speed is the absolute value of velocity. Velocity can be positive or negative. Speed is always positive. Speed is the magnitude of velocity. Speed is the scalar quantity. Velocity is a vector. Vectors have magnitude and direction. Scalar quantities only have magnitude only. Now if you notice, because the acceleration is negative, the velocity is always decreasing. On the left side, it goes from 30 to 0, so it's decreasing. On the right side, from 0 to negative 30, it's still decreasing. So the velocity is always decreasing. However, the speed, because it's never negative, it's positive on the left side and the right side. So notice that the speed is decreasing on the left side, but on the right side it's increasing. because it goes from 0 to 30. So whenever the ball travels upward, the velocity decreases and the speed decreases. But as the ball begins to fall back down to the ground, or towards the earth, the speed increases and the velocity continues to decrease. Make sure you understand that. Now, there's three types of trajectories that you need to be familiar with. So, here's the first one. Let's say if we have a ball that rolls off a cliff and it falls down. H represents the height of the cliff. R is the range, which is the horizontal distance between the base of the cliff, which is here, and where the ball lands. Now, because it's moving horizontally initially, the initial speed is vx and not v. Well, v and vx are the same, but the initial speed is vx. At the top, whenever it's moving horizontally, vy is equal to 0, as in the case of the other problem. So you can easily find the height using this equation, h equals 1 half at squared. It comes from this equation. Now because height is a vertical displacement, everything in this equation that we use has to be in the y direction. So we need to use dy, vertical displacement, vy initial, times t, plus 1 half. a y t squared now dy is basically the height at the top vy is 0 so this term disappears and so you get this equation so make sure you know this equation h equals 1 half a t squared is very useful for this type of trajectory Now if you wish to calculate the range, it's equal to vx team. Remember we said that whenever an object moves with constant speed, d is equal to v team. Now vx is v cosine theta, and vy is v sine theta. It's based on this triangle. Here's v, vx, and here's vy, and here's the angle theta. Now, according to SOHCAHTOOR, perhaps you've seen this if you've taken trig, S stands for sine. sine theta is equal to O&H stands for opposite and hypotenuse so sine theta is the ratio between the side that's opposite to theta which is Vy divided by the sine that's the hypotenuse which is across the box and that's V cosine theta is equal to the adjacent side Vx divided by the hypotenuse of V tangent theta is equal to the opposite side, which is Vy, divided by the adjacent side, Vx. So from this equation, if you multiply both sides by V, you can see that Vy is V sine theta. Now for the second equation, If you do the same thing, Vx is V sine theta. Now for the third equation, it's useful to calculate the angle. If tangent theta is Vy over Vx, then the angle theta is the inverse tangent of Vy divided by Vx. Now according to the Pythagorean theorem, there's one more equation that we can write. Since Vx, Vy, and V are all the three sides in the right triangle, and according to the Pythagorean theorem, C squared equals A squared plus B squared. V squared is equal to V. x squared plus Vy squared. Therefore, V is the square root of Vx squared plus Vy squared. If you need to find the final speed just before it hits the ground, this is the equation that you'll need. Now, let's talk about the second trajectory that you'll see, which occurs when a ball is kicked off from the ground. It goes up and then goes back down. This is going to be the height, or the maximum height of the trajectory, and the horizontal distance is the range. Let's call this point A, B, and point C. Now, if we need to find the time it takes... to go from point A to point B, it's equal to V sine theta divided by G. In another video, I show you how you can derive these equations. Now the time it takes to go from... Let's say a to c is simply twice the value from a to b due to the symmetry of the graph. So you get this equation. Now the maximum height is equal to v squared sine squared theta divided by 2g. The range is equal to v squared sine 2 theta divided by g. So because of the velocity, the ball is kicked at an angle. You're going to be given v instead of vx or vy. And the angle relative to the horizontal is theta. So typically, in a problem like this, you'll be given the initial speed and the angle. And you can find anything you need based on these equations. Now the other equations still apply. Now it's important to understand that the speed at part A, or at point A, is the same as the speed at point C. If you recall, when I went over the trajectory that looks like this, with all the speeds every second, the speed one second later was equal to the speed that occurred at the same height. I think in the last example, one second later, the vertical speed was 20, and on the right side, the vertical velocity was negative 20, which means the speed was also 20. Two seconds later, we had 10 and 10. But as you can see, Whenever you have two points at the same height, the speed will be the same. So if you need to find the final speed at point C, it's the same as the initial speed at point A. Just keep in mind, at point A, the vertical velocity is positive, but at point C, it's negative, but the magnitudes are the same. Now the last trajectory that you need to be familiar with looks like this. Typically you have a ball at a cliff or at a building, and it's kicked off at an angle. It goes up, and then it goes back down. So all of the equations that you've seen before, you can apply it to this trajectory. Now, in addition to the other equations, one equation that might be useful is this equation. y final equals y initial plus vy initial t plus 1 half at squared or ay t squared where ay y is negative 9.8. y initial is basically the height of the cliff. And if you want to find the time it takes to hit the ground, that is at point C, y final replace with zero and then use the quadratic equation to get the answer. Now typically you'll be given the velocity v and the angle theta. So to find vy, vy is v sine theta. So just keep that in mind. Now, to use the quadratic formula, make sure you have a 0 on one side of the equation. t is going to be equal to negative b plus or minus the square root b squared minus 4ac divided by 2a. Now, if you don't want to use the quadratic formula to find the time it takes to go from a to c, you can find the time it takes to go from a to c by using the quadratic formula. to B and add it from B to C. The time it takes to go from A to B, we have that formula. It's going to be 2V sine theta. Actually, we found the 2. It's V sine theta over G. And to find the time it takes to go from B to C, use this equation. You may need to find the height between A and B. But if you look at the right side of the graph between B and C, you can use the fact that H... plus y initial, that's the total height relative to the ground, is equal to 1 half a t squared. And if you solve for t, that's going to give you the time it takes to go from b to c. Then you add up these two values, and you should get this answer. Now, another thing that you may need to know is you need to calculate the speed just before it hits the ground. So, Vx is constant. It's going to be the same at point A, B, and C. However, Vy changes. So, first, you need to find Vy final at point C, using Vy initial at point A. Once you... have Vy final and you already know Vx, you could find Vx by using this equation. You can calculate the final speed of the ball just before it hits the ground using this formula. And to find the angle, use this. It's inverse tangent Vy divided by Vx. To find the maximum height, or the height between point A and B, you can use this equation. It's V squared, sine squared, over 2G. Now let's work on some problems. A ball rolls horizontally off a cliff at 20 meters per second. It takes 10 seconds for it to hit the ground. Calculate the height of the cliff and the horizontal distance traveled by the ball. Now the first thing we need to decide is which of the three common trajectories do we have. So the ball starts off on a cliff, and it rolls horizontally, so it's going to the right, and eventually it's going to fall down. So this is the type of trajectory that we have in this problem. Next, make a list of what you have. We have the speed of the ball as it leaves the cliff. Since it's traveling horizontally, we have the speed vx. So vx is equal to 20 meters per second. Now it takes 10 seconds for it to hit the ground, so we have the time. Calculate the height of the cliff, that's h, and the horizontal distance traveled by the ball, which is the range. To find the height of the cliff, we could simply use this equation, h is equal to 1 half at squared. Now, A, we can use 9.8. If you use negative 9.8, H is negative, but what it really means is that the vertical displacement is negative because the ball is going down. But, for all practical purposes... we want the height to be positive so let's plug in positive 9.8 and t is 10 half of 9.8 is 4.9 10 squared is 100 so 4.9 times 100 is 490. So that's the height of the cliff since it takes 10 seconds to hit the ground. Now let's calculate the range. The range is equal to Vx T and we know Vx is V cosine theta but we already have Vx which is 20. So therefore the range is going to be 20 meters per second multiplied by 10 seconds and you can see that the unit seconds will cancel. 20 times 10 is 200. So the range is 200 meters. And that's it for this problem. A ball rolls off a cliff that is 200 meters high. Calculate the time it takes for the ball to hit the ground. So for this problem, We have a similar trajectory. It rolls off the cliff and then hits the ground. And our goal is to find the height. And we know that the height is 1 half a t squared. Actually we have the height. Our goal is to find the time it takes to hit the ground. The height is 200. The acceleration is 9.8. Now let's solve for t. So half of 9.8 is 4.9. To isolate t squared, we need to divide both sides by 4.9. 200 divided by 4.9 is about 40.8163, and that's equal to t squared. So now we need to take the square root of both sides to get t. And so T is going to be 6.389 seconds. And so that's how long it's going to take for the ball to hit the ground. A ball is released from rest and drops straight down starting at a height of 800 meters. How long will it take to hit the ground? Now what if the ball was thrown straight down with initial speed of 30 meters per second? How long will it take to hit the ground now? So let's draw a picture. So let's say this is the ground level. And here we have a ball. And we're throwing it down. Well, for the first one, it's going to be released from rest. And it's going to fall straight down. What is the initial speed if it's released from rest? Now there's no Vx component because it's not moving to the left or to the right, it's falling straight down. So initially, Vy is equal to 0. If it simply, if it drops down and being released from rest. Now, in the other example, Vy is not 0. The ball was thrown straight down, so vy is negative 30. The speed is positive, but the velocity is negative because it's going in the negative y direction. In each case, the height is 800 meters. So what can we do to find the time it takes for it to hit the ground? In the first example, you can simply use the equation h is equal to 1 half a t squared. h is 800. The acceleration is 9.8. And we can solve for t. Half of 9.8 is 4.9. 800 divided by 4.9. that's 163.265 and that's equal to t squared so now we can take the square root of both sides to get t so t is 12.78 seconds Now what about part B? How can we find the time it takes for it to hit the ground if there's an initial speed? So recall that this equation comes from this expression. dy is equal to vy initial multiplied by t plus 1 half ay times t squared. Now, we need to be careful with the way we are going to use this equation. We're going to put the negative signs in the problem, so we have to be careful. Using this equation, we didn't have to worry about the negative signs. Now the ball is traveling 800 meters in the negative y direction, so the vertical displacement is negative 800. We do have an initial velocity in the y direction, it's negative 30, so we need to take that into account. And then we have plus 1 half. Ay, the acceleration in the y direction, is negative 9.8. So this is going to be negative 4.9 t squared. Now let's move everything from the right side to the left side. So this is going to be positive 4.9 t squared on the left side. plus 30t, and the 800 is going to stay on the left side, so it's still negative 800. As you can see, we have a quadratic equation in standard form. To solve for t, we need to use the quadratic formula. So let's make some space first. so here's the quadratic formula t is equal to negative b plus or minus square root b squared minus 4ac divided by 2a so a is 4.9 9b is 30 C is negative 800 so this is going to be negative 30 plus or minus square root 30 squared is 900 and then this is going to be minus 4 times a which is 4.9 times t squared I mean times C which is negative 800 divided by 2a or 2 times 4.9 which is 9.8 so now let's multiply these three numbers negative 4 times 4.9 times negative 800 that's equal to positive 15,680 we still have a 900 next to it so now let's add those two numbers If we add 900, it's going to be 16,580, and the square root of that number is 128.76. Now, there could be two answers. Negative 30 minus 128.76 and plus 128.76. We know time can't be negative, so we're just going to ignore the negative answer. So, if we take negative 30 plus 128.76, that's positive 98.76, and divide it by 9.8. This will give us a t-value of 10.08 seconds, which is less than the other answer, which was 12.78. Because in the second example, the ball was thrown towards the ground, it's going to take a shorter time to get to the ground, since it was given that initial speed. And so it makes sense why it would be less. Now we can check the answer. If you plug it into the original equation, we need to make sure that we get 0. 0. 4.9 times 10.08 squared plus 30 times 10.08, that's 800.27 minus 800, that's approximately 0. So this answer is correct.

Displacement is equal to velocity multiplied by the time. Now under constant acceleration, displacement is equal to the average velocity multiplied by time. And the average velocity is basically the average of the initial and the final velocity. You add up the initial and the final and you divide it by 2 if the acceleration is constant.

Now you also have some other equations. V final is equal to V initial plus AT. So the final velocity is equal to the initial velocity plus the acceleration multiplied by the time. Also, the square of the final velocity is equal to the square of the initial velocity plus 2 times the product of the acceleration and the displacement.

And there's another equation. Displacement is equal to V initial t plus 1 half a t squared. Now, the displacement is the change in position.

It can be the final position minus the initial position along the x-axis, or it can be along the y-axis. So if we replace D with Y final minus Y initial, we can get another equation that perhaps you've seen in your physics course by now, which looks like this. The final position is equal to the initial position plus V initial T plus 1 half A T squared.

Now, D can represent displacement, or sometimes you can use it to find distance. Distance and displacement are the same if the object moves in one direction and doesn't change direction. Whenever the object changes direction, displacement and distance are different, so you've got to be careful in order to solve it.

But if it's moving in one direction, distance and displacement, they're the same in that case. Now, let's say if we have a ball, and we're going to kick it off the ground, it goes up, and then it's going to go down. Gravitational acceleration is negative 9.8 meters per second squared. For this example, we're going to round it, and we're going to use negative 10, just to keep things simple.

So, let's say at t equals 0, I'm going to put the time inside the ball. Let's say the horizontal velocity, Vx, we're going to say it's 7, and the vertical velocity, Vy, we're going to say it's 30. Now, one second later, what do you think the horizontal and the vertical velocity will be? What about two seconds later, and then three seconds later, and so forth? So one second later, I should probably put that in a different color, the horizontal velocity will not change.

It's going to remain 7. The vertical velocity will change. And how much will the vertical velocity change by? The vertical velocity changes by the gravitational acceleration. Acceleration tells you how much the velocity changes every second.

Since the gravitational acceleration, which is the acceleration in the y direction, it changes about 10. Every second the Vy is going to decrease by 10 every second so one second later. It's going to be 20 2 seconds later the Vx is going to be the same, but Vy is going to be 10 3 seconds later Vy is now 0 at the top. It's not going up anymore.

You only have horizontal motion, but Vx is still 7 Then Vy becomes negative But Vx will always remain 7. Because Vx is constant, the acceleration in the x direction is 0. A projectile, by definition, is only under the influence of gravity. So if you throw a pen, once you release it from your hand, it's a projectile. A flying bird is not a projectile. It's in the air, but it has the force generated by its wings acting on it.

So it's not a projectile. Projectiles can only have gravity acting on it. Air resistance is ignored.

So this is going to be 2, 3, 4 seconds later, Vy is going to be negative 10. 5 seconds later, negative 20. 6 seconds later, negative 30. Now, here's a question for you. What is the vertical velocity and the speed 5 seconds later? The vertical velocity, Vy, is negative 20. The speed is positive 20. Speed is the absolute value of velocity.

Velocity can be positive or negative. Speed is always positive. Speed is the magnitude of velocity.

Speed is the scalar quantity. Velocity is a vector. Vectors have magnitude and direction.

Scalar quantities only have magnitude only. Now if you notice, because the acceleration is negative, the velocity is always decreasing. On the left side, it goes from 30 to 0, so it's decreasing.

On the right side, from 0 to negative 30, it's still decreasing. So the velocity is always decreasing. However, the speed, because it's never negative, it's positive on the left side and the right side. So notice that the speed is decreasing on the left side, but on the right side it's increasing.

because it goes from 0 to 30. So whenever the ball travels upward, the velocity decreases and the speed decreases. But as the ball begins to fall back down to the ground, or towards the earth, the speed increases and the velocity continues to decrease. Make sure you understand that.

Now, there's three types of trajectories that you need to be familiar with. So, here's the first one. Let's say if we have a ball that rolls off a cliff and it falls down. H represents the height of the cliff. R is the range, which is the horizontal distance between the base of the cliff, which is here, and where the ball lands.

Now, because it's moving horizontally initially, the initial speed is vx and not v. Well, v and vx are the same, but the initial speed is vx. At the top, whenever it's moving horizontally, vy is equal to 0, as in the case of the other problem. So you can easily find the height using this equation, h equals 1 half at squared.

It comes from this equation. Now because height is a vertical displacement, everything in this equation that we use has to be in the y direction. So we need to use dy, vertical displacement, vy initial, times t, plus 1 half. a y t squared now dy is basically the height at the top vy is 0 so this term disappears and so you get this equation so make sure you know this equation h equals 1 half a t squared is very useful for this type of trajectory Now if you wish to calculate the range, it's equal to vx team.

Remember we said that whenever an object moves with constant speed, d is equal to v team. Now vx is v cosine theta, and vy is v sine theta. It's based on this triangle.

Here's v, vx, and here's vy, and here's the angle theta. Now, according to SOHCAHTOOR, perhaps you've seen this if you've taken trig, S stands for sine. sine theta is equal to O&H stands for opposite and hypotenuse so sine theta is the ratio between the side that's opposite to theta which is Vy divided by the sine that's the hypotenuse which is across the box and that's V cosine theta is equal to the adjacent side Vx divided by the hypotenuse of V tangent theta is equal to the opposite side, which is Vy, divided by the adjacent side, Vx.

So from this equation, if you multiply both sides by V, you can see that Vy is V sine theta. Now for the second equation, If you do the same thing, Vx is V sine theta. Now for the third equation, it's useful to calculate the angle. If tangent theta is Vy over Vx, then the angle theta is the inverse tangent of Vy divided by Vx. Now according to the Pythagorean theorem, there's one more equation that we can write.

Since Vx, Vy, and V are all the three sides in the right triangle, and according to the Pythagorean theorem, C squared equals A squared plus B squared. V squared is equal to V. x squared plus Vy squared. Therefore, V is the square root of Vx squared plus Vy squared. If you need to find the final speed just before it hits the ground, this is the equation that you'll need.

Now, let's talk about the second trajectory that you'll see, which occurs when a ball is kicked off from the ground. It goes up and then goes back down. This is going to be the height, or the maximum height of the trajectory, and the horizontal distance is the range.

Let's call this point A, B, and point C. Now, if we need to find the time it takes... to go from point A to point B, it's equal to V sine theta divided by G. In another video, I show you how you can derive these equations.

Now the time it takes to go from... Let's say a to c is simply twice the value from a to b due to the symmetry of the graph. So you get this equation. Now the maximum height is equal to v squared sine squared theta divided by 2g.

The range is equal to v squared sine 2 theta divided by g. So because of the velocity, the ball is kicked at an angle. You're going to be given v instead of vx or vy.

And the angle relative to the horizontal is theta. So typically, in a problem like this, you'll be given the initial speed and the angle. And you can find anything you need based on these equations.

Now the other equations still apply. Now it's important to understand that the speed at part A, or at point A, is the same as the speed at point C. If you recall, when I went over the trajectory that looks like this, with all the speeds every second, the speed one second later was equal to the speed that occurred at the same height.

I think in the last example, one second later, the vertical speed was 20, and on the right side, the vertical velocity was negative 20, which means the speed was also 20. Two seconds later, we had 10 and 10. But as you can see, Whenever you have two points at the same height, the speed will be the same. So if you need to find the final speed at point C, it's the same as the initial speed at point A. Just keep in mind, at point A, the vertical velocity is positive, but at point C, it's negative, but the magnitudes are the same. Now the last trajectory that you need to be familiar with looks like this. Typically you have a ball at a cliff or at a building, and it's kicked off at an angle.

It goes up, and then it goes back down. So all of the equations that you've seen before, you can apply it to this trajectory. Now, in addition to the other equations, one equation that might be useful is this equation. y final equals y initial plus vy initial t plus 1 half at squared or ay t squared where ay y is negative 9.8.

y initial is basically the height of the cliff. And if you want to find the time it takes to hit the ground, that is at point C, y final replace with zero and then use the quadratic equation to get the answer. Now typically you'll be given the velocity v and the angle theta. So to find vy, vy is v sine theta.

So just keep that in mind. Now, to use the quadratic formula, make sure you have a 0 on one side of the equation. t is going to be equal to negative b plus or minus the square root b squared minus 4ac divided by 2a. Now, if you don't want to use the quadratic formula to find the time it takes to go from a to c, you can find the time it takes to go from a to c by using the quadratic formula.

to B and add it from B to C. The time it takes to go from A to B, we have that formula. It's going to be 2V sine theta.

Actually, we found the 2. It's V sine theta over G. And to find the time it takes to go from B to C, use this equation. You may need to find the height between A and B.

But if you look at the right side of the graph between B and C, you can use the fact that H... plus y initial, that's the total height relative to the ground, is equal to 1 half a t squared. And if you solve for t, that's going to give you the time it takes to go from b to c.

Then you add up these two values, and you should get this answer. Now, another thing that you may need to know is you need to calculate the speed just before it hits the ground. So, Vx is constant.

It's going to be the same at point A, B, and C. However, Vy changes. So, first, you need to find Vy final at point C, using Vy initial at point A.

Once you... have Vy final and you already know Vx, you could find Vx by using this equation. You can calculate the final speed of the ball just before it hits the ground using this formula.

And to find the angle, use this. It's inverse tangent Vy divided by Vx. To find the maximum height, or the height between point A and B, you can use this equation. It's V squared, sine squared, over 2G.

Now let's work on some problems. A ball rolls horizontally off a cliff at 20 meters per second. It takes 10 seconds for it to hit the ground. Calculate the height of the cliff and the horizontal distance traveled by the ball.

Now the first thing we need to decide is which of the three common trajectories do we have. So the ball starts off on a cliff, and it rolls horizontally, so it's going to the right, and eventually it's going to fall down. So this is the type of trajectory that we have in this problem. Next, make a list of what you have.

We have the speed of the ball as it leaves the cliff. Since it's traveling horizontally, we have the speed vx. So vx is equal to 20 meters per second. Now it takes 10 seconds for it to hit the ground, so we have the time.

Calculate the height of the cliff, that's h, and the horizontal distance traveled by the ball, which is the range. To find the height of the cliff, we could simply use this equation, h is equal to 1 half at squared. Now, A, we can use 9.8.

If you use negative 9.8, H is negative, but what it really means is that the vertical displacement is negative because the ball is going down. But, for all practical purposes... we want the height to be positive so let's plug in positive 9.8 and t is 10 half of 9.8 is 4.9 10 squared is 100 so 4.9 times 100 is 490. So that's the height of the cliff since it takes 10 seconds to hit the ground. Now let's calculate the range.

The range is equal to Vx T and we know Vx is V cosine theta but we already have Vx which is 20. So therefore the range is going to be 20 meters per second multiplied by 10 seconds and you can see that the unit seconds will cancel. 20 times 10 is 200. So the range is 200 meters. And that's it for this problem.

A ball rolls off a cliff that is 200 meters high. Calculate the time it takes for the ball to hit the ground. So for this problem, We have a similar trajectory.

It rolls off the cliff and then hits the ground. And our goal is to find the height. And we know that the height is 1 half a t squared. Actually we have the height. Our goal is to find the time it takes to hit the ground.

The height is 200. The acceleration is 9.8. Now let's solve for t. So half of 9.8 is 4.9.

To isolate t squared, we need to divide both sides by 4.9. 200 divided by 4.9 is about 40.8163, and that's equal to t squared. So now we need to take the square root of both sides to get t.

And so T is going to be 6.389 seconds. And so that's how long it's going to take for the ball to hit the ground. A ball is released from rest and drops straight down starting at a height of 800 meters.

How long will it take to hit the ground? Now what if the ball was thrown straight down with initial speed of 30 meters per second? How long will it take to hit the ground now? So let's draw a picture. So let's say this is the ground level.

And here we have a ball. And we're throwing it down. Well, for the first one, it's going to be released from rest.

And it's going to fall straight down. What is the initial speed if it's released from rest? Now there's no Vx component because it's not moving to the left or to the right, it's falling straight down.

So initially, Vy is equal to 0. If it simply, if it drops down and being released from rest. Now, in the other example, Vy is not 0. The ball was thrown straight down, so vy is negative 30. The speed is positive, but the velocity is negative because it's going in the negative y direction. In each case, the height is 800 meters. So what can we do to find the time it takes for it to hit the ground?

In the first example, you can simply use the equation h is equal to 1 half a t squared. h is 800. The acceleration is 9.8. And we can solve for t.

Half of 9.8 is 4.9. 800 divided by 4.9. that's 163.265 and that's equal to t squared so now we can take the square root of both sides to get t so t is 12.78 seconds Now what about part B? How can we find the time it takes for it to hit the ground if there's an initial speed?

So recall that this equation comes from this expression. dy is equal to vy initial multiplied by t plus 1 half ay times t squared. Now, we need to be careful with the way we are going to use this equation. We're going to put the negative signs in the problem, so we have to be careful.

Using this equation, we didn't have to worry about the negative signs. Now the ball is traveling 800 meters in the negative y direction, so the vertical displacement is negative 800. We do have an initial velocity in the y direction, it's negative 30, so we need to take that into account. And then we have plus 1 half. Ay, the acceleration in the y direction, is negative 9.8.

So this is going to be negative 4.9 t squared. Now let's move everything from the right side to the left side. So this is going to be positive 4.9 t squared on the left side.

plus 30t, and the 800 is going to stay on the left side, so it's still negative 800. As you can see, we have a quadratic equation in standard form. To solve for t, we need to use the quadratic formula. So let's make some space first. so here's the quadratic formula t is equal to negative b plus or minus square root b squared minus 4ac divided by 2a so a is 4.9 9b is 30 C is negative 800 so this is going to be negative 30 plus or minus square root 30 squared is 900 and then this is going to be minus 4 times a which is 4.9 times t squared I mean times C which is negative 800 divided by 2a or 2 times 4.9 which is 9.8 so now let's multiply these three numbers negative 4 times 4.9 times negative 800 that's equal to positive 15,680 we still have a 900 next to it so now let's add those two numbers If we add 900, it's going to be 16,580, and the square root of that number is 128.76.

Now, there could be two answers. Negative 30 minus 128.76 and plus 128.76. We know time can't be negative, so we're just going to ignore the negative answer. So, if we take negative 30 plus 128.76, that's positive 98.76, and divide it by 9.8.

This will give us a t-value of 10.08 seconds, which is less than the other answer, which was 12.78. Because in the second example, the ball was thrown towards the ground, it's going to take a shorter time to get to the ground, since it was given that initial speed. And so it makes sense why it would be less.

Now we can check the answer. If you plug it into the original equation, we need to make sure that we get 0. 0. 4.9 times 10.08 squared plus 30 times 10.08, that's 800.27 minus 800, that's approximately 0. So this answer is correct.

Transcript for:Understanding Projectile Motion Fundamentals

Transcript for:
Understanding Projectile Motion Fundamentals