A ball is dropped from a height of 36 ft. How long does it take the ball to hit the ground.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

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.

Approved by eNotes Editorial Team
An illustration of the letter 'A' in a speech bubbles

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.

Approved by eNotes Editorial Team