A 500 g model rocket is on a cart that is rolling to the right at a speed of . The rocket engine, when it is fired, exerts an 8.0 N vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the ground. At what horizontal distance left of the hoop should you launch?

Video transcript

Hi, everyone in this particular practice problem, we are asked to determine the horizontal distance left of the rim where the athletes should release the ball to achieve the target where a basketball player was running at a constant speed of four m per second along the positive X axis while holding a 0.625 kg ball, the player actually releases the ball by applying a vertical force of eight Newton on the ball. And the players target is to actually enter the ball through the basket rim from below the ball and the rim are in the same xy plane and the vertical distance separating the player's hand and the rim is two m. So we want to determine the horizontal distance left of the rim where the athletes should release the ball to actually enter it to the basket. So the options are 0.29 m, 2.6, 3.5 and 4.6 m. So we will model the ball as a particle like object. And in this problem, I am going to use the subscript B for the ball and are for the rim. And we will consider the release point. Our position as X B Not equals YB not affordable equals zero m at time, T equals zero seconds. So X B not and YB not equals zero is essentially the players happens. So the ram position with respect to this is that you're going to be X R. And why are, and why are is going to equals to two m With respect to the release point here, which is given in the promised statement, which is the vertical distance separating the player's hand and the Rim is two m. So the ball's horizontal velocity Phoebe X is going to be plus four m per second constant, which is the speed that the basketball player is running at. So the ball's vertical acceleration is still unknown. And the time t when the ball enters the rim from below is also unknown. So those two are the two unknowns that we have to actually calculate first in order for us to determine the distance, the horizontal distance left of the rim where the athletes should release the ball, which is essentially X R right here, which is the one that we are looking for. So first, we want to calculate the acceleration, we're calculating the acceleration by applying Newton's second law along the vertical distance. So I'm just gonna write down Newton's second law on the vertical access or direction. So vertically um there are two different forces acting upon the ball. So first, we have the actual weight of the ball obviously oops. So we have the weight of the ball which is M multiplied by G and we also have the upward force that the player is releasing, which is F um F B for a ball which is 88 Newton. So in the vertical direction, F net Y in the Y axis is equals to M multiplied by A B Y for the acceleration of the ball in the Y direction the F net is just F B minus uh W equals M multiplied by A B Y. So we are looking for A B Y, so I'm just gonna rearrange this a little bit. So A B Y is equals to F B and W is, is just multiplied by G and then define everything by another M. So F B is eight Newton, we know all these values here from the problem statement. So FB is eight Newton M is 0.6 kg, which is the mass of the ball, the G normal 9.81 m per second squared. And the other M is also um 0.625 kg just like. So and this will give us an A B Y value or acceleration of the ball in the vertical direction value of 2. m per second squared just like. So, so next, after we find the A B Y, the second step is to actually find the time P when the ball enters the rim. So when the ball enters the rim, we won the Y R equals Y B, Which is essentially equals to two m. So now finding the time the winnable understeer, um we can use the Kinnah Matic equation Y R equals Y B not plus fee B Y I multiplied by T plus half A B Y T squared. So this is just essentially the distance, which is why equals the original or initial distance. Why be not? Which is essentially zero. So we can cross this odd or neglect this plus fee I initial T plus half A B Y acceleration T squared. So the Y R is going to be two m, oops two m equals Phoebe Y I which is the acceleration in the Y direction is also initially zero T plus half A B Y 2.99 m per second squared T squared just like. So, so this right here can also be neglected equals to zero. So essentially the T can be calculated by, so T is essentially the square root of two multiplied by two divided by uh 2.99 just like. So, So with this, the T is going to be 1.16 seconds. So at 1.16 seconds, the ball actually enters the Rim. Thus the time T is the time for the ball to move up and then enter the room. Now that we find the time that it takes for the ball to actually enter the rim, we can calculate the horizontal distance from the ball to the point of the release, oh, from the rim to the point of the release of the ball, which is the athletes or the player's hand. So finding the horizontal distance, we can use X -1 or essentially what we're looking for is just X R um X one minus X F equals X R that will equals to Phoebe X multiplied by T. So recall that the Phoebe access constant at four m per second and the T is the time that it takes for the ball to reach the rim, which is 1.16 seconds, the one that we just found. So X are essentially is going to come up before 10. m and 4.6 m is going to be the distance where the ball should be released from to the left of the rim. So that will be the answer to this particular problem. So 4.6 m is going to essentially be option D and option D is going to be the answer to our problem Where the horizontal distance left off the Rim where the athletes should release the ball to achieve the target is going to be 4.6 m. And that will be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be it. Thank you.