When Babe Ruth hit a homer over the 7.5-m-high right-field fence 95 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 38 degree angle with the ground.

Respuesta :

95m /2 = 47.5m
 
47.5m = Vocos38t

Vy = Voy -9.8t

0 = Vosin38 - 9.8t
 0 = Vosin38 - 9.8(47.5 / Vocos38)
0 = Vo^2(0.62) - 590.7

Vo = 30.87 m/s

Hope this helps