When baseball outfielders throw the ball, they usually allow it to take one boun

Question

When baseball outfielders throw the ball, they usually allow it to take one bouncer, 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 the figure below), but loses half its speed. (a) Assuming 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 (blue path) as a ball thrown upward at phi - 45.0 degree with no bounce (green path)? (b) Determine the ratio of the times for the one-bounce and no-bounce throws.

Explanation / Answer

D =ucosx(2u sin)/g

= u2 sin2/g

ii)

the distance before bounce be d1 and after bounce (d-d1)

such that they add up to D

d1 = u2sin2/g

(D-d1) =  u2sin2/4g [velocity becomes half, such that u' = u/2]

D =5u2sin2/ 4g

Equating both:

5u2sin2/ 4g = u2 sin2/g

= 26.56 ° Answer

(b)

T1B = 2usin/g

T0B = 2usin/g

T1B / T0B = 0.447/ 0.707

= 0.632 Answer

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