A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releasing it at the bottom of the circle (underhand pitch). If the speed of the ball at the top of the circle was been 15 m/s, what will be the speed just after it's released?

I need some help! Please show the equation and then plug in the numbers into that equation

3 answers

Ke up top = 1/2*.25 *225
gain of Pot energy = .25*9.81*1.2
work input = (1/4)(2 pi *.6)*30

so
sum of those 3 energies =
(1/2)(.25)v^2
Thank you so much!
Why the 1.2?