The general falling body equation is usually given as:
`S=-16t^2+v_0t+s_0` where t is measured in seconds and the distances are measured in feet. `v_0` is the initial velocity -- here it is zero as the ball is dropped from rest. `s_0=36` since the initial height is 36ft. (The -16 is an approximation for the earth's gravitational pull.)
Then letting S=0 we get:
We ignore the negative solution.
Thus the ball hits the ground 1.5 seconds after being dropped.
We can use the equations of motion for this question.
`S = Ut+1/2*a*t^2`
`S = 36ft`
U = 0 since ball is dropped freely
`a = 32.174ft/s^2`
`36 = 0+1/2*32.174*t^2`
` t = 1.496s`
So the ball will drop to ground after 1.496 seconds once it dropped.