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

a) y = 103.21 m

b) t = 4.6 s

Step-by-step 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) = (V_(f)^(2))/(2g) = ((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 = -(V_(f))/(g) = -(- 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!