Two part acceleration and force question

Question

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 68 mph. The 0.20 kg ball is 55 cm from the pivot point at her shoulder.

Just before the ball leaves her hand, what is its centripetal acceleration in m/s^2? answer in two significant figures

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? Answer in Newtons and two significant figures

Answer 1

( 68 mph ) ( 1609 m / mile ) ( 1 hour / 3600 s ) = 30 m/s.

centripetal acceleration = v^2 / r = ( 30 m/s )^2 / ( .55 m )^2 = 1600 m/s^2

F = m a = ( .20 kg ) ( 1600 m/s^2 ) = 330 N