- #1

Lukapaka

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## Homework Statement

When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle as it had when released but loses half its speed.

1) Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce (blue path) as one thrown upward at α = 49.8° with no bounce (green path)?

2) Determine the ratio of the times for the one bounce and no bounce throws.

## Homework Equations

1) d = v x t

2) Vh = V x cos(theta)

## The Attempt at a Solution

d1 = distance of throw without bounce

d2 = distance of throw with one bounce

d2 = da + db = distance traveled before bounce + distance traveled after bounce

d1 = V x (cos(49.8)) x t

d2 = V x (cos(alpha)) x ta + (V/2)x(cos(alpha)) x tb

d1 = d2

V x (cos(49.8)) x t = V x (cos(alpha)) x ta + (V/2)x(cos(alpha)) x tb

V cancels and I'm left with:

cos(49.8) x t = (cos(alpha)) x ta + (1/2)x(cos(alpha)) x tb

I can't get the times to cancel out, and don't know any other way to approach this question. Help!