Question

How long does it take for a ball dropped off of a building of height 52.9 m to hit the ground if the ball starts its fall from rest?

Answer 1

Given:

Height of the building where the ball is dropped

`h=52.9\text{ m}`

it is given that the ball starts from rest which means the initial velocity of the ball is zero.

`u=0\text{ m/s}`

Required:

how long does it take for a ball to reach the ground

`t=?`

Step-by-step explanation:

When a ball is dropped at some height above the ground then the ball is starts coming down to earth with some acceleration which is the acceleration due to gravity.

then to solve this problem we use the third equation of motion which is given as

`h=ut+(1)/(2)gt^2`

here, h is the height, and t is the time, g is the acceleration due to gravity that equal to

`9.8\text{ m/s}^2`

plugging all the values in the above relation, we get

`\begin{gathered} h=ut+(1)/(2)gt^2 \\ 52.9\text{ m}=0* t+(1)/(2)*9.8\text{ m/s}^2* t^2 \\ t^2=\frac{52.9\text{ m}*2}{9.8\text{ m/s}^2} \\ t^2=10.79\text{ s} \\ t=\sqrt[2]{10.79} \\ t\approx3.28\text{ s} \end{gathered}`

Thus, the time taken by a ball to reach the ground is

`3.28\text{ s}`