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? Assume the ball wa hit 1.0 m above the ground and it path initially made a 38 degree angle with the ground.

Answer 1

Dh = Vo^2*sin(2A)/g = 95 m.
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.

Answer 2

2.2 m/s