I\'m trying to solve the following problem... A football is punted with a horizo
ID: 1725343 • Letter: I
Question
I'm trying to solve the following problem...A football is punted with a horizontal components of velocity of 35m/s. It has a hang time of 4.6 seconds. Find thehorizontal range of the football, its initial vertical component ofvelocity and its initial angle of projection.
I know that the horizontal range is 35 m/s x 4.6 s = 161 m but explain the following to me
which of the following solutions is correct and what's wrong withthe reasoning of the other...
since 4.6 s = total time, 2.3 s = time up (0 - Vi,up) /-9.8 m/s2 = 2.3 seconds and V i, up (initial verticalcomponent of velocity) = 9.8 x 2.3 = 22.5 m/s and tan-1 = 22.5/35 ... and = 33 degrees
since 4.6 x = total time, 2.3 s = time up (0 - Vi x sin 35)/-9.8 m/s2 = 2.3 seconds and sin = 0.644 making = 40 degrees....
What am I missing? Be basic and complete in your answer,please!
Explanation / Answer
=33 degrees. you could have verified that by findingthe initial velocity (35=V0*cos33), and then plugging itback in to find the hang time(-4.9t2+V0*t*sin33=0, and I got t=4.63seconds). I can't quite tell what you're trying to do in the second part, butit looks like you're trying to use the fact that the verticalvelocity is zero at the peak of the trajectory, by knowing that thetime was 2.3 seconds, and the vertical velocity is zero. butyou didn't find either the initial velocity or the initialangle. I don't know why you would assume thatsin=.644(=22.5/35?). the reasoning was correct in thefirst part, where you assumed thatsin-1()=22.5/35.