a pitched ball is hit by a batter at a 45 degree angle and it just clears the outfield fence, 110m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.

Answer 1

the height y of the ball at distance x is

y = x tanθ - g/2v^2 sec^2 θ x^2

so, we want to solve for v in

110 - 9.8/v^2 * 110^2 = 0
v = 32.83 m/s