When babe ruth hit a homer over the 10 m -high right-field fence 80 m from home plate, roughly what was the minimum speed of the ball when it left the bat? assume the ball was hit 0.90 m above the ground and its path initially made a 45 ∘ angle with the ground

Refer to the diagram show below.

Let V =  the speed of the ball when it left the bat.
The horizontal component of velocity is
Vx = Vcos(45°) = 0.7071V
The vertical component of velocity is
Vy = V sin(45°) = 0.7071V

Assume g = 9.8 m/s² and ignore air resistance.

The time of flight is
t = (80 m)/(Vx m/s) = 80/(0.7071V) = 113.1382/V s

The difference in height between the launch point and the fence is 
h = 10 - 0.9 = 9.1 m.
Therefore



29.7 m/s (nearest tenth)

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n