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Carnegie Mellon University

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Problem 58 Hard Difficulty

A football receiver running straight downfield at 5.50 $\mathrm{m} / \mathrm{s}$ is 10.0 $\mathrm{m}$ in front of the quarterback when a pass is thrown downfield at $25.0^{\circ}$ above the horizon (Fig. P3.58). If the receiver never changes speed and the ball is caught at the same height from which it was thrown, find (a) the foothall's initial speed, (b) the amount of time the football spends in the air, and (c) the distance between the quarterback and the receiver when the catch is made.

Answer

(a) $n _ { 0 } = 14.5 \mathrm { m } \mathrm { s } ^ { - 1 }$
(b) $t = 1.25 \mathrm { s }$
(c) $R = 16.43 \mathrm { m }$

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Video Transcript

So we first know that for part a time is gonna be equaling two velocity initial sign of data over G. And in this case, Fada is gonna be equaling 25 degrees. So this would be the time that the football is spending in the air. So this would be equaling to the initial sign of 25 degrees divided by G. So we can then say that the horizontal range of the football are is gonna be equaling 10 meters, plus the velocity times the t the time that the football is in the air. The left side accounts for the horizontal range of the football and then the right side accounts for the distance or the displacement of the receiver. And so the receiver is moving at a constant velocity, no acceleration. So we don't have to account for that term. And initially, it is 10 meters ahead of the football, right when the football is launched. So we can then say that of the initial squared sine of tooth ada divided by G using their range equation for the range of the football, this is gonna be equaling 10 meters, plus the constant velocity of the receiver V multiplied by T that we had just found. And so we can then substitute all of this in and say that then for party V initial squared sine of two times 25 degrees. So two times 25 degrees and then this would be divided by 9.8 meters per second squared. This would equal 10 meters. This would be a plus 5.50 meters per second, multiplied by T that we had just found to the initial sign of 25 degrees divided by 9.8 meters per second squared. And you can use your T 84 85 or 89 in order to solve for the initial and the initial is found to be equaling 1.47 meters per second. So for part A, this would be the initial velocity of the football. In order for the receiver to perfectly catch the football. Now we can solve really quickly. If the time that the footballs in the air this would be two times the initial of 1.47 meters per second, multiplied by sine of 25 degrees, divided by 9.8 meters per second squared and the time that the footballs in the air is 0.1 to 9 seconds. So this would be the time that the football is in the air and we can then say that the distance between the quarterback and the receiver is a lot easier to find. This would be the distance this would be equaling 10 meters plus 5.50 meters per second, multiplied by 0.1 to 9 seconds and the distances that eve willing 10 0.7 one meters. This would be the distance between the quarterback and the receiver as the receiver catches the football after the quarterback has thrown it 0.1 to 9 seconds beforehand. That is the end of the solution. Thank you for watching.

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