Question

A ball is held at rest and then dropped. The final velocity of the ball was 45 m/s just before it hit the ground. How high was the ball when it was dropped?How much time past before the ball hit the ground?

Answer 1

Answer:

a) y = 103.21 m

b) t = 4.6 s

Explanation:

a) The height at which the balls was is given by:

$V_{f}^{2} = V_{0}^{2} - 2g(y_{f} - y_{0})$

Where:

$y_{f}$: is the final height = 0

$y_{0}$: is the initial height =?

$V_{f}$: is the final velocity = 45 m/s

$V_{0}$: is the initial velocity = 0 (is held at rest

g: is the gravity = 9.81 m/s²          

Hence, the height is:

$y_{0} = \frac{V_{f}^{2}}{2g} = \frac{(45 m/s)^{2}}{2*9.81 m/s^{2}} = 103.21 m$

b) The time before the ball hit the ground is:

$V_{f} = V_{0} - g*t$

$t = -\frac{V_{f}}{g} = -\frac{- 45 m/s}{9.81 m/s^{2}} = 4.6 s$

Therefore, 4.6 seconds passed before the ball hit the ground.

I hope it helps you!