Question

An object is projected into the air with a vertical velocity of 20 feet per second. At what times will the object be on the ground

Answer 1

20 ft/sec / 32 ft/sec^2 = .625 to reach top

2 * .625 = 1.25

It will be on the ground at t = 0 and t = 1.25

You can also solve the equation:

H = 0 = V0 t - 1/2 g t^2

Obviously, at t=0 h = 0

V0 = 1/2 g t is a solution

t = 20 * 2 / 32 = 1.25 sec