- #1
nickhassan38
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Homework Statement
A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.4 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 15.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?
Homework Equations
KE = 1/2mv^2
PE = mgh
The Attempt at a Solution
1/2mv^2 = 1/2mv^2 + mgh
225.125 = .125v^2 +26.4
198.725 = .125v^2
7.43