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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?
13 years ago

Answers

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
13 years ago
Answered by Anonymous
46.578224
8 years ago

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