Hey, guys. So up until now, all the projectile motion problems that we've seen so far, we've had projectiles launched from stationary positions, like when you are standing somewhere and you throw an object horizontally or downwards. However, you're going to see some problems in which projectiles are launched from vehicles. Those vehicles are already moving with some velocity. For example, like a plane that's releasing a package while it's in mid-air. So what we want to do first is actually distinguish those two objects. We're going to refer to the object that gets dropped or released as the projectile, and the object from which it gets released is going to be the vehicle. So when we talk about the velocity of that vehicle, we're going to denote this as \( v_{\text{vehicle}} \). So I'm going to show you in this video how to solve problems where you're releasing or launching projectiles from moving vehicles. It really just comes down to a straightforward equation that you will be able to use all of the same steps that we've been using for projectile motion. So let's check it out.
In this first example here where you have an object or a projectile that's launched from a moving airplane, the airplane's moving along at 300 meters per second, and it releases the package. So what happens to this package here? Well, if it's just launched or released, the way I like to think about this is that the velocity of this projectile basically just borrows the velocity of the vehicle that it was released from, which is 300 meters per second. So if this package here just gets released and the plane's moving at 300 meters per second, then the projectile also will be moving at 300 meters per second. And so eventually, it's just going to take this parabolic path because once it's released, it is subjected only to the influence of gravity, and it's going to behave just like any other horizontal launch problem.
Let's take a look at another example where you have a balloon that's moving downwards. You are moving downwards at 3 meters per second, and you're going to launch an object outwards at 4 meters per second. So what is the velocity of this projectile here? Well, you're launching it at 4 meters per second outwards. But just like the other example, it's going to borrow the 3 meters per second downwards of the balloon that it was just launched from. So what happens is these two velocities actually combine together, and they're going to produce a 2-dimensional downward launch velocity or downward launch problem like this. So what is the velocity of this projectile? Is it 4 meters per second? Well, no. Because again, it's going to borrow some of the 3 meters per second. And so what I have to do is we actually have to add these 2 velocity vectors together, and so this \( v_{\text{projectile}} \) is going to be 5. Remember, this is just going to behave just like a vector addition problem. So we've got these two components here, and the 2-dimensional velocity vector we get using the Pythagorean theorem.
Let's take a look at this last example here. We've got a vehicle that's moving at 30 meters per second and you launch an object straight upwards. So what happens is this projectile is not going to go upwards or to the right, but actually a combination of those two things. And then it's just going to take a normal parabolic path just like any other symmetrical launch type problem. So what is the projectile? Well, you're launching it at 40 meters per second, but it's borrowing some of the 30 meters per second, of the car that it was just launched from. So we have to use the Pythagorean theorem just like we did over here. So this is going to be my triangle like this. And so my projectile is going to be at 50 because this is a 30, 40, 50 triangle.
So, that's really, you know, how you combine these velocities together. So really what we can see here is there's actually just a simple equation to figure out \( v_{\text{projectile}} \). It's just going to be the velocity of the launch plus the velocity of the vehicle that it was launched from. But these are vectors here. So \( v_{\text{projectile}} \) is actually just the vector addition of the Launch Velocity and the velocity that it borrows from the moving vehicle. And remember, the last thing here is that in the situation, the special case, where you're just dropping or releasing an object, what that means is that \( V_{\text{launch}} \) is just equal to 0. So if you just have a projectile that simply dropped or released, then the moving vehicle and the projectile just move at the same velocity. So that's why this projectile was 300 and the vehicle was also 300. That's really it for this one, guys.
So, basically what happens is that in order to use all your known system of steps and equations that we've been using so far, sometimes you just have to figure out the velocity of the projectile as a first step, and then everything else is going to be the same. So let's take a look at this example here. We've got a cart that's carrying a vertical missile launcher. The cart moves along at a speed of 60 meters per second to the right. So that's going to be the vehicle. Right? So this vehicle here, \( v_{\text{vehicle}} \) is 60, and we have this missile launcher that's going to go upwards, and we have \( v_{\text{launch}} \) is equal to 80. So really what happens is that these two velocities will combine, and and you'll have this projectile that actually moves like this. And so this \( V_{\text{projectile}} \) here is going to be the Pythagorean theorem of \( 60^2 \) and \( 80^2 \). So this is just going to be 100, and then this projectile is just going to take a normal parabolic path just like any otherricular_symmetrical launch problem. So all we have to do now is just stick to the steps. We're going to draw the path in the x and y. This is just going to look exactly like this. And then in the Y axis, it's just going to look like this. It's going to go up and down. And so our points of interest are a and then the maximum height of B and then back down to its original height c. So this is going to behave like any other symmetrical launch problem. We're just going to use the exact same steps, right? So let's go ahead and figure out what we're looking for our target variable. Well, we're looking for the maximum height achieved by the rocket. So for step 2, we're looking for delta y. So, which interval are we going to look at? Well, if we're looking at the interval, if we're trying to figure out the maximum height here from the ground, then we're just going to look at the interval from A all the way up to the maximum height, which is B. So we're going to be looking at the interval A to B. And so our target variable is delta y from A to B. And so we've got our \( A_{y} \) equals negative 9.8. Our \( V_{initialY} \) is \( V_{AY} \). Our final \( Y \) is \( V_{BY} \), and then we've got delta y from A to B. That's what we're looking for. And then we've got time, \( t_{AB} \). So what is what is our \( V_{AY} \)? Well, this is just going to be the the way that we would normally solve this is we would take this \( V_{\text{projectile}} \) here and the angle, and then we would decompose it into its x and y components. But the thing about these kinds of problems is that we actually already know what the y component of the velocity is. It's just \( V_{\text{launch}} \) because we know that this missile is just launched upwards at 80 meters per second. So this is really just \( V_{\text{launch}} \) here, and we know this is equal to 80. And so we also know that \( V_{BY} \) is equal to 0. Remember that these, within these symmetrical launch problems, the velocity here at the peak is equals to 0. So this just behaves like any other normal symmetrical launch problem. And then what we can use is we can just ignore this variable here, and so we can use equation number. That's going to be 2. And so \( V_{BY}^{2} \) equals \( V_{AY}^{2} \) plus \( 2A_{Y} \) times delta y from A to B. We know this is 0. So this is going to this is going to be \( 80^2 \) plus \( 2 \times -9.8 \times \delta y \) from A to B. And so if you go ahead and work this out, you're going to get delta y from A to B is equal to 326.5 meters. And that's the answer. That's the maximum height achieved by the rocket. So again, the first step is just to figure out this initial projectile velocity, and then you're going to stick to all the normal steps that you've seen before. That's it for this one, guys. Let me know if you have any questions.
