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...

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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:

`-16t^2+36=0`

`-16t^2=-36`

`t^2=9/4`

`t=+-3/2`

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.*