Consider this problem. A ball rolls off from an 80 meter cliff and lands on the ground 120 meters away from the base of the cliff. What is the initial velocity of the ball? In this video, we're going to focus on problems like this, finding the initial velocity given a projectile motion problem.
So let's begin by drawing a picture. So here is the ball. It's going to roll off a cliff and it's going to follow that typical trajectory.
Our goal is to calculate the initial velocity. We know the height of the cliff. It's 80 meters.
And we also know that the ball lands 120 meters away from the base of the cliff. So this is the range, and this is the height. So given the range and the height, what formula would you use to calculate the initial velocity? For those of you who want a direct formula, here it is.
The initial velocity is equal to the range times the square root of the gravitational acceleration divided by 2h. Now you might be wondering, well, how'd you get this formula? Well, you could find this formula in my formula sheet. I have a lot of derived equations that will help you go straight to the answer for those of you who want that.
So I'm going to put the link in the description section below to that formula sheet for those of you who want to print it out. But let's talk about how we can get this formula. So first, perhaps you've seen this equation.
Vertical displacement is equal to V sine T. minus 1 half gt squared. Vertical displacement dy is y final minus y initial. So sometimes you might see this equation as y final is equal to y initial plus v sine theta t minus one half gt squared or sometimes you'll see plus one half at squared. For projectile motion the acceleration in the y direction is negative so if you see this negative sign here you want to plug in positive 9.8 for g.
If you see this equation in this form, plus 1 half a t squared, you want to plug in negative 9.8 for a. This term overall should be negative when dealing with projectile motion, if you want to get the right answer. Now, let's focus on V initial.
Notice that it's horizontal, which means that it doesn't have a Y component. It's completely in the X direction. So V initial is the same as VX.
At the top of the trajectory, When it's moving in a horizontal direction, Vy is 0. So for this problem, Vy initial is 0. This part here, V initial sine, that's Vy, or Vy initial. So this part here, V initial sine theta, that's 0. While it's horizontal, the angle is 0 degrees. Sine of 0 degrees will give you 0. So we can get rid of this part. So we get dy is equal to negative one half gt squared.
dy is basically the vertical displacement. And going to the bottom of the trajectory or the ground, dy will equal h. The only difference is dy is negative, h is positive.
So h is the absolute value of dy, so h is going to be positive 1 half gt squared. dy can be positive or negative, it's a vector based on direction, but the height of the building will always be designated a positive value. Now the other equation that we could use is this, the range is equal to vxt.
Now, because Vy initial is 0, V initial is equal to Vx. So the range, we could say, is just V initial T. Now, solving for V initial in this equation, if I divide both sides by T, I get V initial is R over T. Or, if I want to write it this way, r times 1 over t.
Now, in this formula, I'm going to solve for t. I'm going to multiply both sides by 2, so I get 2h is equal to gt squared. 1 half and 2 will cancel.
And then I'm going to divide. by g. So I get 2h over g is equal to t squared. Take the square root.
So we get that t is equal to the square root of 2h over g. But here we need 1 over t. So if t is equal to that, 1 over t is simply the reciprocal of what we have here. So it's going to be g over 2h. So now I can replace 1 over t with this.
And when I do that, I'm going to get this equation. r times the square root of g over 2h. So anytime you have a problem like this with this exact trajectory, if you need to calculate the initial velocity, all you need is the range and the height of the building.
And then with this formula, you can calculate the initial velocity. So let's go ahead and complete the problem. So the range is 120. G is 9.8 and the height of the building is 80. So 120 times the square root of 9.8 divided by 160. That's 21 square root 2. which is 29.7 meters per second. So that is the initial velocity of the ball as it rolls off the cliff. Now let's move on to our second problem.
So a water hose shoots water from the ground at an angle of 30 degrees above the horizontal. Let's draw a picture. So let's say this is the water hose.
And it shoots water. And it follows that trajectory. Now we know the angle is 30 degrees.
The stream of water lands 8 meters away. So that tells us the range of the projectile. In this case the projectile is a fluid. What is the initial velocity of the water as it exits the holes? So once again, we want to calculate the initial.
So we don't know the height. There's no building. So we can't use the previous equation.
So how can we calculate the initial velocity given just r and theta? Well, if you have access to my formula sheet, you can identify this particular formula for this type of trajectory. V initial is equal to the square root of Rg divided by sine 2 theta.
So with that formula, you can get the initial velocity directly. But let's talk about how we can derive that formula. So we know the range is equal to vxt, and vx is the initial cosine theta.
Now, the time it takes to go from point A to point C Basically, the time of flight. This equation is also in the formula sheet. It's 2V sine theta over G.
Going from A to B is just V sine theta over G. But from A to C, it's 2V sine theta over G. Now, what I'm going to do is I'm going to replace T with what we have here.
So r is equal to v initial cosine theta, by the way this is also v initial, times 2 v initial sine theta over g. So now we have r is equal to, we have two v initials, so multiplying those two will give us the initial squared. Now it's helpful to know the double angle formula of sine.
Sine 2 theta is 2 sine theta cosine theta. So here we have 2 sine theta cosine theta. We can replace that with sine 2 theta.
If you access my formula, sheet in Patreon, I have another formula sheet for trig. So in that trig formula sheet, you'll see the double angle formula for sine 2 theta. So in that same formula sheet for projectile motion, I have this formula.
If you want to calculate the range, all you need is the initial velocity and the angle. Now what we're going to do is we're going to rearrange that formula. to get the one above, to get this.
So let's solve for V initial. First, I'm going to multiply both sides by G. So I'm going to have RG, this is going to cancel, is equal to V initial squared, sine 2 theta.
Next, I'm going to divide by sine 2 theta on both sides. So I get Rg over sine 2 theta is equal to the initial squared. And then we could take the square root to get the initial, and that's how we get that equation.
Now, let's go ahead and finish this problem. So R is going to be 8, G is 9.8, and then sine 2 theta. Theta is 30, so 2 theta will be 60. So it's going to be the square root of 8 times 9.8 divided by sine 60. Once you know the formula, you can get the answer very quickly. So the initial speed of the water leaving the hose is 9.51 meters per second. So that is the initial speed.
The initial velocity is speed with direction, so you would say 9.51 meters per second at an angle of 30 degrees above the horizontal. For the first problem, we had 29.7 meters per second. If you want to put an angle to it, it's at 0 degrees, since it's horizontal for number 1. Number three.
A ball is launched at an angle above the horizontal from a 100 meter tall building and lands on the ground 300 meters away from the base of the building 10 seconds later. What is the initial launch angle of the ball? So we have this building. And we have a ball. It's launched at an angle and then it falls down.
We don't know the launch angle, nor do we know the initial velocity. We need to find those two. We do know the height of the building.
It's 100 meters, so we have h and we know the range. it landed 300 meters away from the ground, so we know r. Let's call this point A, the top point B, and when it hits the ground, point C.
So how can we determine the initial launch angle of the ball? Well, if you have the formula sheet, which you can find in the links below, you could use this formula to get the launch angle. It's R tangent, 1 half, gt squared, minus y initial, divided by R.
With this, all you need is three things. You need the range, which we have. You need y initial, which is basically the height of the cliff, and the time. We know that it's going to reach the ground 10 seconds later. So we have everything we need to find this answer.
So at point C, T is 10 seconds. So you can find this equation and all of the other equations in the formula sheet. Now the formula sheet contains way more equations than what you're seeing today.
A lot more. Probably equations you haven't seen before with projectile motion, but once you take a look at it It'll make your life easier because you could find just one equation that will give you the answer directly If you were to pause this video, if I didn't share this equation with you, how would you solve this problem? Before I show you how to derive this equation, if you didn't know this equation existed... What would you do to solve it? Because the previous two problems, you were only missing, you were only looking for one thing.
But for this problem, we have two unknown variables. The initial velocity and the angle theta. What equation would you have used besides this one to get the answer? Feel free to pause the video and see if you could find a way to solve this problem if you didn't know this equation existed.
You can unpause the video when you're ready to see the answer. So let's go ahead and begin. Let's talk about how we can derive this equation.
So let's start with this. y final is equal to y initial plus v initial sine theta t minus one half gt squared. Now, we're going from point A to point B.
At point A, y initial is 100. It's the height of the cliff. At point B, y final, not point B, but point C, y final is 0 because it's at ground level. So, we could put a 0 here. Now, what I'm going to do for this equation is I want to isolate the initial. I want to solve for the initial velocity.
So y final is 0. I'm just going to scratch that off. This term, I'm going to move it to the other side. It's negative on the right side, so it's going to be positive on the left side. So on the left side, we're going to have 1 half gt squared.
y initial, I'm also going to move to the left side. It's positive on the right, but it will be negative on the left. So this is what I now have.
Now to get V initial by itself, I'm going to divide both sides by sine T, or T sine theta. So I'm going to rewrite the equation here. V initial is 1 half gT squared minus y initial over T sine theta. If we have the angle...
We have the height of the cliff, we have the time it takes for it to hit the ground. If we knew the angle, we could use this formula to get the initial velocity of the ball. But we don't know the angle yet, so we can't do that right now.
Now, there's another equation that we can use. Since we have two missing variables, we need two equations to solve those two missing variables. The other one is this.
Range is equal to vx t, and we know vx is v initial cosine theta t. So in this equation, I'm going to solve for v initial. All I have to do is divide both sides by cosine t, or t cosine.
So I get that V initial is equal to R over T cosine theta. Now, because these two equations equal to the initial, that means that they equal to each other. So let's set them equal to each other.
1 half gt squared minus y initial over t sine theta is equal to r over t cosine theta. Now what I'm going to do is I'm going to multiply both sides by t sine theta. So on the left, these two will cancel.
On the right, t will cancel. So I get one half. gt squared minus y initial is equal to r, and then sine theta over cosine theta. If you have the trig formula sheet, you know that's tangent theta, unless you already took trig.
Now, I'm going to isolate theta, so I'm going to divide by r. So I get that tan theta. is one half gt squared minus y initial over r.
Now to get the angle theta, I need to take the arc tangent. both sides our tan and tan will cancel which will give me theta on the left and On the right is what we have here. So that's how you could derive that formula So the formula sheet has a lot of derived equations That will help you get the answer directly You know for those of you who need help solving projectile motion problems So it's a tool to help you keep track of the equations you need to know.
But it's good to understand the process of solving these problems and rearranging equations. So we have 1 half times g, which we're going to plug in positive 9.8. And then t is 10 seconds.
minus y initial, which is the height of the building, so that's 100, over r. r is 300. So 10 squared is 100 times 1 half, that's 50. 50 times 9.8 is 490. Minus 100, that's 390. 390 divided by 300 is 1.3. So we're going to take the arc tangent of 1.3.
And that will give us the initial launch angle of 52.43 degrees. Alright, so now that we have the initial launch angle, let's determine the initial velocity of the ball. Now we had two equations in which we can do that, and we wrote them down earlier.
The first one, well actually the second one is a lot easier. The initial is r t cosine theta. The other equation that we could use is the initial is one half g t squared minus y initial.
over t sine theta. Both of these equations are on the formula sheet. So if you need to find initial velocity and if you know the range and the angle, you can use that formula. If you know the height of the building, the angle, and the time it takes for the projectile to hit the ground, you could use this one to get the answer.
But in both cases though, you need the launch angle to find the initial velocity for this type of problem. Let's use the easier equation. So the range is 300, the time is 10 seconds, and then it's going to be cosine 52.43.
300 divided by 10 is 30. 30 divided by cosine 52.43 gives us an initial velocity of 49.2 meters per second. So if this problem asked you for the initial speed, this will be your answer. 49.2 meters per second.
But since we want the velocity, velocity includes both the magnitude and the direction. So the answer will be... The initial velocity is 49.2 meters per second, launched at an angle of 52.4 degrees.
So you got to mention the angle for velocity as well. You need to give the direction in addition to the magnitude. So that's basically it for this video. By the way, for those of you who want more videos or more example problems on projectile motion, feel free to check out the links in the description section below. If you want access to just more videos on physics, like projectile motion, vectors, kinematics, free-fall problems, motion time graphs like velocity time graphs, forces, you know, Newton's first second third law check out my website video-tutor.net you can access my physics video playlist there along with my other playlists on calculus algebra trig pre-cal physics chemistry organic chemistry you can get access to all those playlists at my website video-tutor.net so feel free to take a look at that when you get a chance