Asked by Yuki
A baseball has a mass of .145 kilograms. A pitcher's fastball travels at 42 m/s. How much force is applied to the ball if the catcher stops the ball in .05 sec?
Answers
Answered by
Henry
a = (Vf - Vo) / t,
a = (0 - 42) / 0.05 = -840m/s^2.
F = ma = 0.145 * (-840) = -122N.
a = (0 - 42) / 0.05 = -840m/s^2.
F = ma = 0.145 * (-840) = -122N.
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