Question

The acceleration due to the Earth's gravity, in English units, is 32 ft/s2. In the absence of air friction, a ball is dropped from rest. Its speed on striking the ground is exactly 132 ft/s. From what height was the ball dropped?

Answer 1

Step-by-step explanation:

You want to first find the amount of time it took using the first kinematic equation.

Vf = Vi + at

-132ft/s = -32ft/s^2(t)

t = 4.125 seconds

Now use the second equation to find the displacement of the ball

Xf = Xi + Vit + 1/2at^2

Xf = 132 + 0(4.125) -16(4.125)^2

Xf = 132 - 16(17.02)

Xf = 132 - 272.32

Xf = -140.32 feet

We arent asked for displacement tho, we were asked for the magnitude of the height. Which is 140.32 feet