Hi I just need confirmation on a problem. Q: A softball pitcher rotates a 0.250-kg ball around a vertical circular path of radius 0.600 m before releasing it. The pitcher exerts a 30.0-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? A: PEg= mgh F=ma; F= 30.0 N, m= .250 kg, a= 30.0 N/.250 kg= 120 m/s^2 v^2= vo^2 + 2ax v^2= (15m/s)^2 + 2*(120m/s^2)*.6pi. v= 26.0 Thanks so much.