when babe ruth hit a homer over the 7.5-m-high right field fence 95 m from home plate, roughly what wa the minimum speed of the ball when it left the bat
Homework Answers
Vo^2*sin(76)/9.8 = 95,
Multiply both sides by 9.8:
Vo^2*sin(76) = 931,
Vo^2 = 931 / sin76 = 959.5,
Vo = 31 m/s.
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when babe ruth hit a homer over the 7.5-m-high right field fence 95 m from home plate, roughly what wa the minimum speed of the ball when it left the bat
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