A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 90 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.

Answer 1

The range of projectile is
L=(v0)^2•sin2α/g.
If α=45 degr, sin2α = 1, then
L=v0^2/g.
v0=sqroot(L•g)= sqroot(90•9.8)= 29.7 m/s