Understanding Projectile Motion Concepts

In this video, we're going to go over some equations that you need to know to solve projectile motion problems. So let's review some basic kinematic equations. Whenever an object is moving with constant speed, displacement is equal to velocity multiplied by time. Now, when an object is moving with constant acceleration, there are four equations. you need to know. The final speed is equal to the initial speed plus the product of the acceleration and the time. There's also this equation. The square of the final speed is equal to to the square of the initial speed plus two times the product of the acceleration and the displacement. Displacement is equal to the average speed, which... The average speed is basically the sum of the initial speed and the final speed divided by 2 multiplied by the time. As you can see, the third equation that I listed here looks like the first equation for a constant speed. Okay, I did not want to do that. So I can rewrite this equation like this. V average times t, where V average is initial plus final divided by 2. Now, there's another equation that you need to know. D is equal to V initial t plus 1 half a t squared. So make sure you know that equation too. So what exactly is D? You can use D as displacement or distance. Distance and displacement are the same if an object moves in one direction. and doesn't change direction. If it changes direction then displacement and distance is not the same. But technically D is displacement. Displacement is basically the difference between the final position minus the initial position. Now you can use displacement in the X direction or you can use it in the Y direction. So just keep that in mind. Now there's three types of shapes for projectile motions that you need to be familiar with. Let's say if we have an object, and it comes off a cliff horizontally and then eventually hits the ground. That's the first type of trajectory you're going to be dealing with. So what equations do we need for this situation? The first equation that's going to help you is this, h is equal to 1 half at squared, where h represents the height of the cliff, and t is the time it takes to hit the ground. This equation comes from this equation, d is equal to v initial t plus 1 half at squared. Now when you're using these equations, you need to ask yourself, am I dealing with the x direction or the y direction? direction you gotta separate x and y for projectile motion problems we're gonna apply this equation along the y-direction so this is really dy is equal to V y initial times t plus 1 half ay times t squared. dy, the displacement in the y direction, is basically the height. So that's h. vy at the top is always 0 because the object is moving horizontally. The initial speed of the ball is vx, not vy. And so we get this equation. So h is equal to 1 half a t squared. So let me just redraw this picture. So make sure you know the first equation that we just mentioned. Now, in addition to knowing this equation, which if you know the time you can find the height of the cliff, there's also another one. We know that d is equal to vt. This is the range of the graph, the distance between where the ball lands and the base of the cliff. Now, if we apply this equation in the x direction, the displacement in the x direction is the range. So, we can say that the range is equal to vxt. And for this type of trajectory, vx is the initial speed of the ball. So, that's how you can calculate the range. If you have the range, you can find the time, and then using the time, you can find the height. So, these are the two main equations that you'll need. for this particular trajectory. Now, there are some other equations. Sometimes you've got to find the speed of the ball just before it hits the ground. Whatever vx was at the beginning, vx would be the same. For projectile motion problems, vx does not change. The acceleration in the x direction is 0, so vx is constant. The acceleration in the y direction, that's the gravitational acceleration, it's negative 9.8. So Vy changes To calculate Vy we can use this equation V final equals V initial Plus 80, but we're going to use it in the y direction So we know that Vy initial is 0 At the top, Vy is always 0. So you can find Vy final using this equation. Now to find the speed of the ball just before it hits the ground, you need to use the horizontal velocity and the vertical velocity. And if you need to find the angle, you can use this equation. It's inverse tangent Vy divided by Vx. Now, the second type of trajectory involves this type of fiction. So, let's say if we have a ball, and we kick the ball from the ground, it goes up, and then it goes down. So that ball is going to have an angle theta relative to the horizontal, and it's going to be launched at a speed v. This is not vx or vy, this is v. Now let's draw the vector v. So here's v, and here's the angle theta. v has an x component and it has a y component. The x component is vx, the y component is vy. So Vx is equal to V cosine theta and Vy is equal to V sine theta. And V, as we mentioned before, is the square root of Vx squared plus Vy squared. And if you want to find theta, it's inverse tangent Vy over Vx. Now, let's define this as point A, point B, and point C. What is the equation that we need to calculate the time it takes to go from A to B, or to reach the maximum height, which is at position B? How long does it take to go from A to B? Now, to derive that equation... We need to start with this equation. V final equals V initial plus AT. But we're going to use it in the Y direction. So this is going to be VY final equals VY initial plus AT. At the top, that is at position B, VY is always 0 because it's not going up anymore. So, B is the final position, A is the initial position. So, VY final is 0. VY initial is basically V sine theta, as we wrote it in this equation. Acceleration in the Y direction is basically G. So solving for t, we need to move this term to the other side. So negative v sine theta is equal to gt. And dividing both sides by g, we can see that t is equal to v sine theta over g. So that's the time it takes to go from a to b. The time it takes to go from... a to c is twice the value. Notice that the graph is symmetrical. So if it takes 5 seconds to go from a to b, it takes 5 seconds to go from b to c. And so to go from a to c, it's 10 seconds. So the total time is just 2v sine theta divided by g. If you're wondering what happened to the negative sign. Know that g is negative 9.8, so it cancels with the negative sign. But if you plug in positive 9.8, you don't need the negative sign anymore. Either case, time is always positive. So just make sure you report a positive value for time. Now, let's talk about some other equations that relate to this shape. How can we derive an expression? to calculate the maximum height between position A and position B. So going from A to B, let's use this particular kinematic equation, V final, or VY final squared. is equal to Vy initial squared plus 2 times Ay times Dy. So you've seen it as V final squared equals V initial squared plus 2Ad. I simply added the subscript y to it because the height is in the y direction. Now at the top, Vy is equal to 0. So, position B is the final position, so Vy final is 0. Vy initial, we know that Vy is V sine theta. So, Vy squared is V sine theta squared. And, of course, we have plus 2 times, the acceleration in the y direction is g, the displacement in the y direction is the same as the height. So, we can put h at this point. So let's move this term to the left. So negative V squared sine squared theta, if you distribute the two, is equal to 2GH. So now let's divide by 2G. So the maximum height, if we ignore the negative sign, is V squared sine squared theta over 2G. So that's how you can find the height if... You have the angle theta and the velocity, not vx, vy, but the velocity v. Now what equation can we come up with in order to calculate the range? How can we derive an equation to find the range of the graph? What equation would you use? The range is the horizontal distance, or the horizontal displacement of the ball. Now we said that the range is equal to vxt, and you can find vx by using the fact that vx is v cosine theta. Now the range is the horizontal displacement from points A to C. So what is the time from point A to C? As we mentioned earlier in this video, the time it takes for the ball to go from A to C is equal to 2V sine theta divided by G. So we have v cosine theta and we're going to replace t with 2v sine theta over g. Now v times v is basically v squared. So we have v squared times 2 sine theta cosine theta divided by g. Now there's something called a double angle formula in trigonometry and here it is. sine 2 theta is equal to 2 sine theta cosine theta. So therefore we can replace the expression 2 sine theta cosine theta with sine 2 theta and so therefore the range is v squared sine 2 theta divided by g. So that's how you can derive an equation for the range if you have this type of trajectory. Now the third type of trajectory that you're going to see involves A ball being launched at an angle from a cliff or from some elevated position above ground level. So it's going to go up and then it's going to go down. So H is the height of the cliff and R, once again, is the range of the ball. Let's call this position A, position B, and position C. So one of the first type of questions that you might find with this type of problem is calculating the time it takes to hit the ground, going from A to C. Now this graph is not the same as this trajectory, where the time it takes to go from A to C is 2V sine theta over G. If you use this equation, that'll give you the time it takes to get to this position, which is symmetrical to A, so don't do it, it's not going to work. So therefore, we need to do something else. It turns out that there's another way to calculate the time. I'm going to show you two ways, so please be patient. Let's start with this equation. Displacement is equal to v initial t plus 1 half a t squared. But we're going to apply it in the y direction. Displacement in the y direction is the difference between the final position minus the initial position. And then, v initial, but in the y direction, is vy initial. Acceleration in the y direction is basically 9.8, or g. So moving this term to the other side, perhaps you've seen this equation. Y final equals Y initial. Y initial is basically the height of the cliff plus VY initial T. Remember, VY initial is V sine theta. Theta is the angle above the horizontal plus 1 half GT squared. Now, to find the time it takes to hit the ground, you need to realize that the position, the y value at ground level is 0. So, if you replace it with 0, replace y initial with h, replace vy initial with v sine theta, you now have everything you need to solve for t. You know what g is. However, you need to use a quadratic equation. You want to make sure everything's on one side and on the other side you have a zero. So using the quadratic formula, t is equal to negative b plus or minus square root b squared minus 4ac divided by 2a. So let's say if you get an equation that looks like this, and you put it in standard form. 4.9t squared. Plus 8t plus 100 or something like that. In this case, well actually this is going to be negative 4.9t squared. You got to make sure you plug in negative 9.8 for g. A is negative 4.9. B is 8. C is 100. And then just plug it into this formula and you should get the answer. Now let's go back to the same trajectory. Now what's another way in which we can calculate the time? That is the time it takes to hit the ground. To go from position A to position C. Is there a way in which we can get the same answer without using the quadratic equation? It turns out that there is. To go from A to B, notice that it's the same as this trajectory. So we can use the equation that we used in that trajectory. The time it takes to go from A to B is simply... v sine theta divided by g. And typically in this problem, you'll be given v. That is just the speed of the ball. And you're going to be given theta, the angle relative to the horizontal. Now how can we find the time it takes to go from b to c? Well, in order to do that, you need to find the height between positions a and position b. To find that height, it's going to be equal to v squared, sine squared, and I believe it was over 2g, if I'm not mistaken. I believe that's the equation. Once you get the height, then the total height, which, let's call the total height y max. The total height from B to C, we can use that to find the time. Now, you've seen this equation, H is equal to 1 half A T squared. But in this particular situation, between B and C, H, it's really the sum of H plus capital H. So that's the total height from B to C. And that's equal to 1 half A, which is the same as G, times... T squared. So basically, if you rearrange the equation, T going from B to C is going to be, let's call this Y max. So 2 times Y max. divided by g, square root. That's going to give you the time it takes to go from B to C. So the total time is simply the sum of these two values. And that's how you can avoid using the quadratic formula if you don't want to. Now, how can we find a range? What equation would you use to calculate the range of the ball? To find the range, use this equation, vx, t. For this trajectory, do not use the equation v squared sine 2 theta over g. Only use this equation if you have a symmetrical trajectory. This is the only time you should use this equation. Otherwise, if you use it for this problem, you're going to get the range between these two points. let's call this point a and point D you get the range between those two points and you don't want that so to find the range for this type of trajectory use the equation vx times T now keep in mind vx is equal to V cosine theta So the range is simply v times cosine theta times t. You can use this form of the equation if you want to. But you need to find the time it takes to go from a to c before you can use it. Which, using the last two equations, you can get that answer now. And that's it. That's all you need to do to find the range. Now sometimes, you might be asked to find the speed of the ball just before it hits the ground. So first, you need to realize that Vx is constant. So whatever Vx you have here, it's going to be the same at point C. At point A, point B, and point C, Vx is the same. So let me make sure I write that. Vx is constant. It does not change. So whatever Vx you have here, we're going to use that to find the final speed of the ball, just before it hits the ground. Now just like last time we mentioned earlier in this video, we've got to find the vertical velocity as well using this equation. So you need to find Vy final, and you know Vy initial is basically V sine theta, where V is the initial velocity at point A. And G, make sure you plug in negative 9.8. T, this is the time it takes to go from A to position C. So use the total time. So this will give you the final vertical velocity. So once you have Vx and Vy at point C, to find the final speed just before it hits the ground, you can now use this equation. And if you need to find the angle, as mentioned before, it's going to be inverse tangent. Vy divided by Vx. Now let's talk about the angle. So let's say this is the ball and it's moving in this direction just before hits the ground. And let's say this is the x-axis and this is the y-axis making the ball the center. So the ball has an x component Vx and it has a y component Vy. You're looking for V, which is the velocity, and the magnitude of velocity is the speed. So once you use inverse tangent theta, make sure you plug in positive values for Vy and Vx. Even though Vy is going to be negative, don't plug in a negative value, plug in a positive value. That will give you the reference angle, which is an angle between 0 and 90. Now sometimes you might describe your answer as being below the horizontal or relative to the positive x-axis. So let's say if the angle, let's say it's 60 degrees. So that's going to be the reference angle. So you can describe it as being 60 degrees below the horizontal, which is the x-axis. Or you could say it's positive 300 relative to the positive x-axis. So you have two ways of describing the angle. Just be careful to get it the right way in terms of the way the problem wants you to describe it. So it could be 60, it could be 300. Just think about what they're asking for. If it's below the horizontal, 60. If it's measured from the positive x-axis, then it's 300. It's 360 minus 60, which is 300. So make sure you understand all of the equations and when to use them. So just to review, for this type of trajectory, where the ball simply falls down, it travels horizontally from a cliff and just falls off, the height is equal to 1 half a t squared, and range is equal to v x t. And for all trajectories, you can use this equation. If you need to find the angle, use this. Just plug in positive values for Vy and Vx. And also, if you need to find Vy final, to use it here, use this equation. Now for the second trajectory, these are the main equations that you need, just to review what we went over. So the time it takes going from, let's say, A to B, remember it's just V sine theta divided by G. And the time it takes to go from A to C is twice that value. It's 2V sine theta over G. And the range... Is V squared, sine squared, divided by, let's see if I remember this, it was, no, that's not the range, that's the height. It was V sine 2 theta, divided by G. That's what it was. And now for the other one, to find the maximum height, it's V squared, sine squared theta, divided by 2G. So those are the four main equations you need for this trajectory. And also, if you need to find the angle just before it hits the ground at position C, it's the same as the angle when it left the ground. And the speed at which it left the ground is the same as the speed at which it hits the ground. So let's say if it left the ground at 20 meters per second, the speed before it hits the ground is 20. Since this trajectory is symmetrical, A and C are the same. The angles can be the same, and the speed will be the same. And then finally, for the last trajectory, all of the equations that you've seen for the first two, you can apply it to this one. The only thing that's different is this equation. Y final equals Y initial plus VY initial T plus 1 half AT squared. And so we're going to stop here. So those are all of the equations that you'll need for these type of projectile motion problems. So that's it for this video, and thanks for watching.

you need to know. The final speed is equal to the initial speed plus the product of the acceleration and the time. There's also this equation. The square of the final speed is equal to to the square of the initial speed plus two times the product of the acceleration and the displacement.

Displacement is equal to the average speed, which... The average speed is basically the sum of the initial speed and the final speed divided by 2 multiplied by the time. As you can see, the third equation that I listed here looks like the first equation for a constant speed.

Okay, I did not want to do that. So I can rewrite this equation like this. V average times t, where V average is initial plus final divided by 2. Now, there's another equation that you need to know. D is equal to V initial t plus 1 half a t squared. So make sure you know that equation too.

So what exactly is D? You can use D as displacement or distance. Distance and displacement are the same if an object moves in one direction. and doesn't change direction.

If it changes direction then displacement and distance is not the same. But technically D is displacement. Displacement is basically the difference between the final position minus the initial position.

Now you can use displacement in the X direction or you can use it in the Y direction. So just keep that in mind. Now there's three types of shapes for projectile motions that you need to be familiar with. Let's say if we have an object, and it comes off a cliff horizontally and then eventually hits the ground.

That's the first type of trajectory you're going to be dealing with. So what equations do we need for this situation? The first equation that's going to help you is this, h is equal to 1 half at squared, where h represents the height of the cliff, and t is the time it takes to hit the ground. This equation comes from this equation, d is equal to v initial t plus 1 half at squared. Now when you're using these equations, you need to ask yourself, am I dealing with the x direction or the y direction?

direction you gotta separate x and y for projectile motion problems we're gonna apply this equation along the y-direction so this is really dy is equal to V y initial times t plus 1 half ay times t squared. dy, the displacement in the y direction, is basically the height. So that's h.

vy at the top is always 0 because the object is moving horizontally. The initial speed of the ball is vx, not vy. And so we get this equation.

So h is equal to 1 half a t squared. So let me just redraw this picture. So make sure you know the first equation that we just mentioned.

Now, in addition to knowing this equation, which if you know the time you can find the height of the cliff, there's also another one. We know that d is equal to vt. This is the range of the graph, the distance between where the ball lands and the base of the cliff. Now, if we apply this equation in the x direction, the displacement in the x direction is the range.

So, we can say that the range is equal to vxt. And for this type of trajectory, vx is the initial speed of the ball. So, that's how you can calculate the range. If you have the range, you can find the time, and then using the time, you can find the height.

So, these are the two main equations that you'll need. for this particular trajectory. Now, there are some other equations. Sometimes you've got to find the speed of the ball just before it hits the ground.

Whatever vx was at the beginning, vx would be the same. For projectile motion problems, vx does not change. The acceleration in the x direction is 0, so vx is constant. The acceleration in the y direction, that's the gravitational acceleration, it's negative 9.8. So Vy changes To calculate Vy we can use this equation V final equals V initial Plus 80, but we're going to use it in the y direction So we know that Vy initial is 0 At the top, Vy is always 0. So you can find Vy final using this equation.

Now to find the speed of the ball just before it hits the ground, you need to use the horizontal velocity and the vertical velocity. And if you need to find the angle, you can use this equation. It's inverse tangent Vy divided by Vx. Now, the second type of trajectory involves this type of fiction. So, let's say if we have a ball, and we kick the ball from the ground, it goes up, and then it goes down.

So that ball is going to have an angle theta relative to the horizontal, and it's going to be launched at a speed v. This is not vx or vy, this is v. Now let's draw the vector v. So here's v, and here's the angle theta. v has an x component and it has a y component. The x component is vx, the y component is vy. So Vx is equal to V cosine theta and Vy is equal to V sine theta. And V, as we mentioned before, is the square root of Vx squared plus Vy squared.

And if you want to find theta, it's inverse tangent Vy over Vx. Now, let's define this as point A, point B, and point C. What is the equation that we need to calculate the time it takes to go from A to B, or to reach the maximum height, which is at position B?

How long does it take to go from A to B? Now, to derive that equation... We need to start with this equation. V final equals V initial plus AT. But we're going to use it in the Y direction.

So this is going to be VY final equals VY initial plus AT. At the top, that is at position B, VY is always 0 because it's not going up anymore. So, B is the final position, A is the initial position. So, VY final is 0. VY initial is basically V sine theta, as we wrote it in this equation.

Acceleration in the Y direction is basically G. So solving for t, we need to move this term to the other side. So negative v sine theta is equal to gt. And dividing both sides by g, we can see that t is equal to v sine theta over g. So that's the time it takes to go from a to b.

The time it takes to go from... a to c is twice the value. Notice that the graph is symmetrical. So if it takes 5 seconds to go from a to b, it takes 5 seconds to go from b to c. And so to go from a to c, it's 10 seconds.

So the total time is just 2v sine theta divided by g. If you're wondering what happened to the negative sign. Know that g is negative 9.8, so it cancels with the negative sign.

But if you plug in positive 9.8, you don't need the negative sign anymore. Either case, time is always positive. So just make sure you report a positive value for time.

Now, let's talk about some other equations that relate to this shape. How can we derive an expression? to calculate the maximum height between position A and position B.

So going from A to B, let's use this particular kinematic equation, V final, or VY final squared. is equal to Vy initial squared plus 2 times Ay times Dy. So you've seen it as V final squared equals V initial squared plus 2Ad. I simply added the subscript y to it because the height is in the y direction. Now at the top, Vy is equal to 0. So, position B is the final position, so Vy final is 0. Vy initial, we know that Vy is V sine theta.

So, Vy squared is V sine theta squared. And, of course, we have plus 2 times, the acceleration in the y direction is g, the displacement in the y direction is the same as the height. So, we can put h at this point. So let's move this term to the left.

So negative V squared sine squared theta, if you distribute the two, is equal to 2GH. So now let's divide by 2G. So the maximum height, if we ignore the negative sign, is V squared sine squared theta over 2G.

So that's how you can find the height if... You have the angle theta and the velocity, not vx, vy, but the velocity v. Now what equation can we come up with in order to calculate the range? How can we derive an equation to find the range of the graph?

What equation would you use? The range is the horizontal distance, or the horizontal displacement of the ball. Now we said that the range is equal to vxt, and you can find vx by using the fact that vx is v cosine theta. Now the range is the horizontal displacement from points A to C.

So what is the time from point A to C? As we mentioned earlier in this video, the time it takes for the ball to go from A to C is equal to 2V sine theta divided by G. So we have v cosine theta and we're going to replace t with 2v sine theta over g. Now v times v is basically v squared.

So we have v squared times 2 sine theta cosine theta divided by g. Now there's something called a double angle formula in trigonometry and here it is. sine 2 theta is equal to 2 sine theta cosine theta. So therefore we can replace the expression 2 sine theta cosine theta with sine 2 theta and so therefore the range is v squared sine 2 theta divided by g.

So that's how you can derive an equation for the range if you have this type of trajectory. Now the third type of trajectory that you're going to see involves A ball being launched at an angle from a cliff or from some elevated position above ground level. So it's going to go up and then it's going to go down.

So H is the height of the cliff and R, once again, is the range of the ball. Let's call this position A, position B, and position C. So one of the first type of questions that you might find with this type of problem is calculating the time it takes to hit the ground, going from A to C.

Now this graph is not the same as this trajectory, where the time it takes to go from A to C is 2V sine theta over G. If you use this equation, that'll give you the time it takes to get to this position, which is symmetrical to A, so don't do it, it's not going to work. So therefore, we need to do something else.

It turns out that there's another way to calculate the time. I'm going to show you two ways, so please be patient. Let's start with this equation.

Displacement is equal to v initial t plus 1 half a t squared. But we're going to apply it in the y direction. Displacement in the y direction is the difference between the final position minus the initial position.

And then, v initial, but in the y direction, is vy initial. Acceleration in the y direction is basically 9.8, or g. So moving this term to the other side, perhaps you've seen this equation. Y final equals Y initial.

Y initial is basically the height of the cliff plus VY initial T. Remember, VY initial is V sine theta. Theta is the angle above the horizontal plus 1 half GT squared. Now, to find the time it takes to hit the ground, you need to realize that the position, the y value at ground level is 0. So, if you replace it with 0, replace y initial with h, replace vy initial with v sine theta, you now have everything you need to solve for t.

You know what g is. However, you need to use a quadratic equation. You want to make sure everything's on one side and on the other side you have a zero. So using the quadratic formula, t is equal to negative b plus or minus square root b squared minus 4ac divided by 2a. So let's say if you get an equation that looks like this, and you put it in standard form.

4.9t squared. Plus 8t plus 100 or something like that. In this case, well actually this is going to be negative 4.9t squared. You got to make sure you plug in negative 9.8 for g. A is negative 4.9.

B is 8. C is 100. And then just plug it into this formula and you should get the answer. Now let's go back to the same trajectory. Now what's another way in which we can calculate the time?

That is the time it takes to hit the ground. To go from position A to position C. Is there a way in which we can get the same answer without using the quadratic equation? It turns out that there is.

To go from A to B, notice that it's the same as this trajectory. So we can use the equation that we used in that trajectory. The time it takes to go from A to B is simply...

v sine theta divided by g. And typically in this problem, you'll be given v. That is just the speed of the ball. And you're going to be given theta, the angle relative to the horizontal. Now how can we find the time it takes to go from b to c? Well, in order to do that, you need to find the height between positions a and position b.

To find that height, it's going to be equal to v squared, sine squared, and I believe it was over 2g, if I'm not mistaken. I believe that's the equation. Once you get the height, then the total height, which, let's call the total height y max.

The total height from B to C, we can use that to find the time. Now, you've seen this equation, H is equal to 1 half A T squared. But in this particular situation, between B and C, H, it's really the sum of H plus capital H. So that's the total height from B to C.

And that's equal to 1 half A, which is the same as G, times... T squared. So basically, if you rearrange the equation, T going from B to C is going to be, let's call this Y max. So 2 times Y max.

divided by g, square root. That's going to give you the time it takes to go from B to C. So the total time is simply the sum of these two values. And that's how you can avoid using the quadratic formula if you don't want to. Now, how can we find a range?

What equation would you use to calculate the range of the ball? To find the range, use this equation, vx, t. For this trajectory, do not use the equation v squared sine 2 theta over g. Only use this equation if you have a symmetrical trajectory.

This is the only time you should use this equation. Otherwise, if you use it for this problem, you're going to get the range between these two points. let's call this point a and point D you get the range between those two points and you don't want that so to find the range for this type of trajectory use the equation vx times T now keep in mind vx is equal to V cosine theta So the range is simply v times cosine theta times t.

You can use this form of the equation if you want to. But you need to find the time it takes to go from a to c before you can use it. Which, using the last two equations, you can get that answer now.

And that's it. That's all you need to do to find the range. Now sometimes, you might be asked to find the speed of the ball just before it hits the ground.

So first, you need to realize that Vx is constant. So whatever Vx you have here, it's going to be the same at point C. At point A, point B, and point C, Vx is the same.

So let me make sure I write that. Vx is constant. It does not change. So whatever Vx you have here, we're going to use that to find the final speed of the ball, just before it hits the ground.

Now just like last time we mentioned earlier in this video, we've got to find the vertical velocity as well using this equation. So you need to find Vy final, and you know Vy initial is basically V sine theta, where V is the initial velocity at point A. And G, make sure you plug in negative 9.8. T, this is the time it takes to go from A to position C.

So use the total time. So this will give you the final vertical velocity. So once you have Vx and Vy at point C, to find the final speed just before it hits the ground, you can now use this equation.

And if you need to find the angle, as mentioned before, it's going to be inverse tangent. Vy divided by Vx. Now let's talk about the angle.

So let's say this is the ball and it's moving in this direction just before hits the ground. And let's say this is the x-axis and this is the y-axis making the ball the center. So the ball has an x component Vx and it has a y component Vy. You're looking for V, which is the velocity, and the magnitude of velocity is the speed.

So once you use inverse tangent theta, make sure you plug in positive values for Vy and Vx. Even though Vy is going to be negative, don't plug in a negative value, plug in a positive value. That will give you the reference angle, which is an angle between 0 and 90. Now sometimes you might describe your answer as being below the horizontal or relative to the positive x-axis. So let's say if the angle, let's say it's 60 degrees.

So that's going to be the reference angle. So you can describe it as being 60 degrees below the horizontal, which is the x-axis. Or you could say it's positive 300 relative to the positive x-axis.

So you have two ways of describing the angle. Just be careful to get it the right way in terms of the way the problem wants you to describe it. So it could be 60, it could be 300. Just think about what they're asking for. If it's below the horizontal, 60. If it's measured from the positive x-axis, then it's 300. It's 360 minus 60, which is 300. So make sure you understand all of the equations and when to use them.

So just to review, for this type of trajectory, where the ball simply falls down, it travels horizontally from a cliff and just falls off, the height is equal to 1 half a t squared, and range is equal to v x t. And for all trajectories, you can use this equation. If you need to find the angle, use this. Just plug in positive values for Vy and Vx. And also, if you need to find Vy final, to use it here, use this equation.

Now for the second trajectory, these are the main equations that you need, just to review what we went over. So the time it takes going from, let's say, A to B, remember it's just V sine theta divided by G. And the time it takes to go from A to C is twice that value. It's 2V sine theta over G.

And the range... Is V squared, sine squared, divided by, let's see if I remember this, it was, no, that's not the range, that's the height. It was V sine 2 theta, divided by G. That's what it was.

And now for the other one, to find the maximum height, it's V squared, sine squared theta, divided by 2G. So those are the four main equations you need for this trajectory. And also, if you need to find the angle just before it hits the ground at position C, it's the same as the angle when it left the ground.

And the speed at which it left the ground is the same as the speed at which it hits the ground. So let's say if it left the ground at 20 meters per second, the speed before it hits the ground is 20. Since this trajectory is symmetrical, A and C are the same. The angles can be the same, and the speed will be the same.

And then finally, for the last trajectory, all of the equations that you've seen for the first two, you can apply it to this one. The only thing that's different is this equation. Y final equals Y initial plus VY initial T plus 1 half AT squared. And so we're going to stop here. So those are all of the equations that you'll need for these type of projectile motion problems.

So that's it for this video, and thanks for watching.

Transcript for:Understanding Projectile Motion Concepts

Transcript for:
Understanding Projectile Motion Concepts