The pitcher throws a pitch from the mound toward home plate 18.4m away. The ball is released horizontally from a height of 1.8m with a velocity of 32m/s. Find the height above ground when it gets there.

Answer 1

time of flight: 18.4/32 sec

hf=hi-4.9t^2 In the vertical
hf=1.8-4.9(t you compute it)^2