A clump of soft clay is thrown horizontally from 9.80 m above the ground with a speed of 20.0 m/s. Assume it sticks in place when it hits the ground At what time will the clay hit the ground

Explanation:

The time to hit the ground will be same as time taken to fall from the height of 9.8 m with initial vertical velocity of zero .

Considering vertical displacement

initial velocity u = 0

displacement s = 9.8 m

acceleration a = g = 9.8 m /s²

time t = ?

s = ut + 1/2 g t²

9.8 = 0 + .5 x 9.8 x t²

t² = 2

t = √2 = 1.4 s

the answer would be 32*f

resonant frequency