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 theta as it had when released (see figure below) but loses half its speed. 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 (lower path) as one thrown upward at alpha = 48.4° with no bounce (upper path)?
******THIS IS NOT WHAT I AM ASKING. I calculated this to be the correct answer of 26.3 degrees; this was only included because it contains information needed to answer the question I am unsure of.
WHAT I AM ASKING: Determine the ratio of the times for the one-bounce and no-bounce throws.
Any of the time equations for projectile motion (unsure what they are)
Dx = Vx(t)
The Attempt at a Solution
I used the above equation and set it up so that the ratio would equal the time of the first divided by the time of the second. Many variables cancelled out here (V's and D's), and I was left an expression ratio = 3cos26.3/cos48.4..this gave me a ratio of 4.05..The computer said this was incorrect. I was told by my classmates that because the question was not exactly clear as to the ratio of what to what, the inverse of the value calculated was correct for some of them. I only have one try left, and am too scared to put it in because I would like to make sure that my derivation of this time ratio equation is correct. Any help would be GREATLY appreciated