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Carnegie Mellon University



Problem 48 Hard Difficulty

When baseball outfielders throw the ball, they usually allow it to take one bounce, on the theory that the ball arrives at its target sooner that way. Suppose that, after the bounce, the ball rebounds at the same angle $\theta$ that it had when it was released (as in Fig. P9.48), but loses half its speed. (a) Assuming that the ball is always thrown with the same initial speed, at what angle $\theta$ should the ball be thrown in order to go the same distance $D$ with one bounce as a ball thrown upward at $45.0^{\circ}$ with no bounce? (b) Determine the ratio of the times for the one-bounce and no-bounce throws.


(a) $\theta = 26.57 ^ { \circ }$
(b) $\frac { T _ { \text {d } } } { 7 _ { \text {b } } } = 1.05$


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Video Transcript

we know that the range equation this is B for part of the range equation is equaling V initial squared sine of to theatre divided by G. And so we have essentially two parts to one one trajectory. So the one trajectory is letting it bounce, setting the baseball bounce. And so we can say that this would be our prime. This would be equaling the initial squared sine of tooth ada over G for the first for the first throw. And then after it bounces, it bounces with half of the same velocity. So we can say that this would be plus the initial over to quantity squared sine of tooth ada over G. So that would be the full range given that it bounces. And then we can say that if it doesn't bounce, are is simply equaling the initial squared over G because we're trying to compare this to an angle where where this angle is 45 degrees. So now, with this angle equaling 45 degrees, we can set these two equal to one another because supposedly this is having the same range. So we can say that then the initial squared over G is going to be equaling the initial squared sign of tooth Ada, divided by G plus the initial squared sine of tooth ada divided by four G. And this is this would essentially describe the trajectory of the ball as it bounces one time with one bounce. And then this would be if it was, if the range which maximized where fada here. Fate as equaling 45 degrees. So this is the maximum range, and then we want to find the angle at which you should throw the ball in order to get to the maximum range. However, we want to get there faster. So we would set these equal to one another. And we can then say that V initial squared over. Chief is equaling the initial squared over G. This would be sign of two fada plus sign of tooth Ada over four. And so we can then say that four part eh, one is going to be equaling Sign of tooth ada, uh, multiplied by one plus 1/4 and this is equaling five over four. Sign of tooth data. And so Fada is gonna be equaling. This would be 1/2 times art sign of four over five. And so this is Seita is equaling 26.6 degrees. So this would be your answer for party. That ball should be thrown at an angle of 26.6 degrees above the horizontal so that it can reached the maximum range, so this would be above horizontal, so ball reaches max range. Uh huh. Fada equals 45 degrees. So this would be your answer for party. And then for part B, we can say the time taken for no bounce would be t And so the time would be to be a sign of Fada over G. Of course here Seita is equaling 45 degrees. In this case, so T is going to be equaling two times the initial velocity over G times sign of 45 degrees. And so this is gonna be equaling. We could say 1.41 times the initial velocity over, Chief. Now, this would be for one bounce for sorry. This would be for no bounce for one bounce. We can say that. Then this would be t prime and t prime is gonna be equaling the time for the first throw freely through the initial throw and then the time after the first bounce. And so we can say that t someone is gonna be equaling to the sign of data over G. And so this would be equaling two to v sign of 26.6 degrees over G. And we can say that then t's of one is equaling. This would be to the over G multiplied by 0.448 Now t's up to is after the second. If that's where the first bounce. So that one bounce. And so this would be two times the initial velocity divided by two. Because only after the first bounce it's only moving with half of its initial velocity times sign again, 26.6 degrees over G. And so we can then say that this would be equaling two 0.448 once again, but without the factor of two. So it's multiplied by the initial developed by G. And so now we can say that T crime is gonna be equaling two chiefs of one plus t's up to, which is simply gonna be equal in 1.34 times the over G or essentially two times 20.448 plus 0.448 So three times 30.448 which is 1.34 multiplied by the initial divided by G And so dividing these to weaken say T prime over T, this would be equal in 1.34 the initial over G divided by 1.41 the initial over G, and we find that four part B T prime over tea is equaling 40.95 So this means that the time it takes for the ball to reach a certain horizontal range is 95% of the time taken if the ball was thrown without any bounce. So this is one bounce, and this is again no bounce. So, as you can see, throwing it at a short of the smaller angle, 26.6 degrees is actually having it, having the baseball get to its destination faster. Then, if you threw it without any bounce, so even though after the bounce it loses initial, it loses its velocity by half, it still gets there faster. It's only 95%. It'll be 95% of the time taken, um, without any bounce so this would be our final answer. So it is true in the sense that baseball players allow the ball base. Probably baseball players allow the ball to bounce once so that it will get to its destination a bit quicker. That is the end of the solution. Thank you for watching.

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