A fast pitch softball player does a \"windmill\" pitch moving her hand through a

Question

A fast pitch softball player does a "windmill" pitch moving her hand through a vertical circular arc to pitch a ball at 67 mph. The 0.18 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. a_c = m/s^2 At the lowest point of the circle the ball has reached its maximum speed What is the magnitude of the hand exerts on the ball at this point? Express your answer using two significant figures. F = N

Explanation / Answer

Centripetal Acceleration is given by:

ac = V^2/r

r = 55 cm = 0.55 m

V = 66 mph = 29.5 m/sec

using given values:

ac = 29.5^2/0.55 = 1582.27 m/sec^2

B.

At lowest point

Fnet = m*a + m*g

Fnet = m*(a + g)

Fnet = 0.18*(1582.27 + 9.81) = 286.57 N

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