Projectiles Launched From Moving Vehicles - Video Tutorials & Practice Problems

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concept

Releasing or Launching Projectiles From Moving Vehicles

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Hey, guys. So up until now all the project on motion problems that we've seen so far we've had projectiles launched from stationary positions like you standing somewhere and you're throwing an object horizontally or downwards or something like that. But you're gonna see some problems 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 we want to do first is we actually want to distinguish those two objects were gonna refer to the object that gets dropped or released as the projectile and the object that it gets released from is gonna be the vehicle. So when we talk about the velocity of that vehicle, we're gonna denote this as V vehicle. So I'm gonna show you this video on 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 would be able to use all of the same steps that we've been using for project emotions. So let's check it out. So in this first example here, where you have an object or projectile that's really launched from a released from a moving airplane here, the airplanes moving along at 300 m 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 off the vehicle that it was released from, which is the 300 m per second. So if this thing if this package here just gets released and the planes moving at 300 m per second and the projectile also have been moving at 300 m per second and so eventually it's just gonna take this parabolic path because once it's released, it's subjected to Onley the influence of gravity, and it's gonna 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, so you're moving downwards at 3 m per second and you're gonna launch an object outwards at 4 m per second. So what is the velocity of this projectile here? Well, you're launching it at 4 m per second outwards, but just like the other example. It's gonna borrow the 3 m per second downwards off the balloon that it was just launched from. So what happens is these two velocities actually combined together and they're gonna produce a two dimensional downward launch velocity or downward launch problem like this. So what is the velocity of this projectile? Is it 4 m per second? Well, no, because again, it's gonna borrow some of the 3 m per second. And so what? I have to dio we actually have to add these two velocity vectors together. And so this v projectile is gonna be five. Remember, this is just gonna be, you know, this is just gonna behave just like a vector addition. Problems. We've got these two components here and the two dimensional velocity vector we get using the Pythagorean theorem. So let's take a look at this last example here. We've got a vehicle that's moving at 30 m per second and you launch an object straight upwards. So what happens is this projectile is not gonna go upwards or to the right, but actually a combination of those two things on. Then it's just gonna 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 m per second, but its borrowing some of the 30 m 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 gonna be my triangle like this. And so my projectile is gonna be at 50 because this is a 30 40 triangle. So s So That's really how you combine these velocities together. So really, we can see here is there's actually just a simple equation to figure out the projectile. It's just gonna be the velocity of the launch, plus the velocity of the vehicle that was launched from and but these air vectors here, So v projectile is actually just the vector addition off the launch velocity and the velocity that it borrows from the moving vehicle. Remember, the last thing here is that in the situation of special case where you're just dropping or releasing an object, what that means is that v launch it's just equal to zero. So if you just have a projectile, it's 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 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 you know the same 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 gonna be the same. So let's take a look at this example here. So we've got a cart that's carrying a vertical missile launcher. The cart moves along at a speed of 60 m per second to the right. So that's gonna be the vehicle, right? So this vehicle here V vehicle is 60 and we have this missile launcher that's gonna go upwards, and we have the launch is equal to 80. So, really, what happens is that these two velocities will combine and you'll have this projectile that actually moves like this. And so this V projectile here is gonna be the Pythagorean theorem of 60 squared and 80 squared. So this is just gonna be 100. And then this project, I was just gonna take a normal parabolic path just like any other symmetrical launch problem. So all we have to do now is just stick to the steps we're gonna draw the past in the X and Y. This is just gonna look exactly like this. And then in the Y Axis is going to look like this is gonna go up and down and so are points of interest are a and then the maximum height of B and then back down to its original height. See? So it's just gonna behave like any other symmetrical launch problem. We're just gonna use the exact same steps, right? So let's go ahead and figure out what we're looking for. Our target variable. We're looking for the maximum height achieved by the rocket. So for step two, we're looking for Delta y. So which interval we're gonna 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 gonna look at the interval from a all the way up to the maximum height which is B So we look 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. RV initial y is va y. Our final y is V b y. And then we've got Delta y from A to B. That's what we're looking for and then we've got time. Ta be. So what is what is our V a y? Well, this is just gonna be the way that we would normally solve. This is we would take this v projectile here and the angle. 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 the launch because we know that this missile is just launched upwards at 80 m per second. So this is really just the launch here. I mean, this is equal to 80 and so we also know that V b Y is equal to zero. Remember that these, uh, in these symmetrical launch problems the velocity here at the peak is equals zero so it's just behaving 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 gonna be too. And so v b y squared equals v A y squared plus to a y times delta y from A to B. We know this is zero. So this is gonna this is gonna be 80 square plus two times negative, 9.8 times negative delta y from A to B. And so, if you go ahead and work this out, you're gonna get Delta Y from A to B is equal to 326.5 m. And that's the answer. That's the maximum height achieved by the rocket. So again, first step is just to figure out this initial projectile velocity, and then you're gonna 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

2

Problem

Problem

A small plane flies horizontally at 20m/s at an altitude of 200m, when it launches a projectile at a speed of 65 m/s at 22.6° below the horizontal. What horizontal distance does the projectile travel before hitting the ground?

A

752 m

B

344 m

C

184 m

D

920 m

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