To warm up for a match, a tennis player hits the 57.0-g ball vert...

Question

To warm up for a match, a tennis player hits the 57.0-g ball vertically with her racket. If the ball is stationary just before it is hit and goes 5.50 m high, what impulse did she impart to it?

Accepted Answer

Hey everyone in this problem, we have a volleyball player and they're going to hit a ball vertically upward during a practice session, it's going to move from stationary to some maximum height. And were asked to determine the impulse imparted on the ball by the player. The first thing we're gonna do is go ahead and just draw out this system. Okay, we're gonna take up as the positive y direction and just do a little diagram of what's going on. Okay, so before the player hits the ball, we have the ball here. Okay, this is a 0.27 kg Ball, K 270g, kg. Okay. And they're going to hit it with their fist and it's gonna cause it to go up. You know when we're thinking about the impulse itself, Okay, We're told that it moves from a stationary position. Okay, So, the initial speed when we're thinking about the impulse, It's going to be zero. All right. What about the final speed? Well, when we're thinking about the final speed with the impulse, we aren't told anything about that. Okay, so we don't know. Now, after some time the ball is going to be at its maximum height. Okay. Maximum height. When it gets to its maximum height, we know that its speed or its velocity will be zero. Okay, it will stop moving momentarily when it gets to that maximum The ball is still .27 kg And we're told that this maximum height is 8.9 m. Okay, so this distance here, all the way down To where the ball first started. Okay, is 8.9 m. Alright, so we're asked to find the impulse imparted on the ball. Okay, so, when we think about impulse J? Well, let's recall that. That's just equal to the change in momentum. Okay, Delta P. Well, what is momentum? Let's recall, momentum is mass times velocity. So, the change in momentum is going to be the mass times the velocity of the impulse final Okay, minus mass times the velocity of the impulse initially. Okay, and this J is what we're looking for. All right, well, we know that initially before the impulse the ball is stationary, so we know B J not is zero. So the second term is going to go to zero. Alright, We know the mass but we don't know this J F V J f k. We don't know the velocity right after that impulse. Okay, and so this quantity we need to find in order to find jay. Okay, well, how can we find that? Well, let's think about our equations of motion. Okay. All right, well, let's consider the velocity of the motion initially. Okay, we don't know we don't know the speed that that ball is moving right after it's been hit. Okay, When it initially starts to move, we don't know what we do know is that this speed is going to be equal to the speed. V J F Okay, so, the final velocity after the impulse is actually going to be the initial velocity that we're thinking about when we're thinking about motion. Okay, so that impulse is just going to impart um momentarily on that bowl to get it moving. And so the final impulse velocity will actually be the initial velocity when we're thinking about it, moving all the way up to its maximum height. Okay, And so this V J. F. Well, this is what we're looking for. We know V. F. This is zero because the ball gets to a maximum height, it will momentarily stop the acceleration while this is acceleration due to gravity negative 9.8 m per second squared. Okay, Delta Y. The change in Y. Well, we know that's 8.9 m. Okay, it goes up 8.9 m and we don't have any information about T. All right, so, we have three variables that we know one that we're looking for. So we can use our um equations and we want to choose the um equation that doesn't include T. Okay, we don't have information about T. And we don't want to find out information about T. So, the equation we're going to use looking at your um equations is V. F. Squared is equal to the not squared, which in this case is V M not squared. Okay, we're talking about the velocity of the motion or the speed of the motion plus two. A delta Y. Okay. Alright, so again, final speed zero. The initial speed of the motion or the initial velocity of that motion. Well, again, that's the final speed of the impart. Sorry, not at the part of the impulse A plus to a negative 9.8 Times Delta Y is 8.9. And our units here we have a meters per second squared, We have delta Y meter. So here we're gonna get meters squared per second squared. Ok, so just rearranging we at V J f squared is equal to 174.44 and again m squared per second squared? We get VJF is equal to plus or 13. 757 m/s. Okay, when we take the square root, we get both positive and negative route and we need to interpret okay, well in this case the motion is going up. We've chosen up to be positive. So we want the positive route. Okay, so we want to take positive route and ignore this negative route. Okay, so now we have our final speed and we can get back to our delta RJ calculation. Okay, so let's go back up to J. So again, we have the change in momentum. The initial speed is zero We have the mass 0.27 kg. Okay, our green quantity this V J F we just found. So times 13.20757 m per second. Now we're going to get an impulse of 3. kg meter per second. Okay, so that is the impulse imparted on the ball. That is going to be answered. That's it for this one. Thanks everyone for watching. See you in the next video.

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