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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?

Respuesta :

Thank you for posting your question here at brainly. Below is the solution:

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

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