Question

A ball is thrown horizontally from the top of a building 96

high. The ball strikes the ground 79 horizontally from the
point of release. What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s

Answer 1

H = 1/2 g t^2 time for ball to fall vertical distance H

t = (2 * 96 / 9.8)^1/2 = 4.43 sec time for ball to fall 96 m

Vy = a t = 9.80 m/s^2 * 4.43 s = 43.4 m/s vertical speed of ball when it reaches the ground

If the ball traveled 79 m horizontally then

Vx =79 m / 4.43 s = 17.8 m/s horizontal speed of ball

V = (Vx^2 + Vy^2)^1/2 total speed of ball

V = (43.4^2 + 17.8^2)^1/2 = 46.9 m/s