Asked by Anonymous

A physics student throws a softball straight up into the air. The ball was in the air for a total of 6.16 s before it was caught at its original position.
(a) What was the initial velocity of the ball?
(b) How high did it rise?
Answered by Damon
time to top = 3.08 s
v = Vi - gt
v is 0 at top
so
Vi = 9.8*3.08

h = Vi t - 4.9 t^2 where t = 3.08
Answered by Anonymous
46.578224
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