Part A A fast pitch softball player does a \"windmill\" pitch, moving her hand t

Question

Part A A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 75 mph. The 0.17 kg ball is 55 cm from the pivot point at her shoulder Just before the ball leaves her hand, what is its centripetal acceleration? Express your answer using two significant figures. ac = m/s Submit My Answers Give Up Part B At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point? Express your answer using two significant figures Submit My Answers Give Up

Explanation / Answer

v = 75.0 mph = 33.52 m/s

m = 0.17 kg

r = 0.55 m

centripetal acceleration = a = v^2/r = (33.52 m/s)^2/(0.55 m) = 2042.9 m/s^2

F = ma = (0.17 kg) (2042.9 m/s^2 ) = 347.3 N

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