Question

A ball is thrown horizontally from the top of a building 22.3 m high. The ball strikes the ground at a point 127 m from the base of the building. The acceleration of gravity is 9.8 m/s^2 . Find the time the ball is in motion. Answer in units of s.

Answer 1

t = 2.13 s

Step-by-step explanation:

given,

height of the building = 22.3 m

horizontal distance = 127 m

acceleration due to gravity = 9.8 m/s²

time for which ball is in motion = ?

using equation of motion

`s = u t + (1)/(2)gt^2`

initial velocity is zero

`s = (1)/(2)gt^2`

`t = \sqrt{(2s)/(g)}`

`t = \sqrt{(2* 22.3)/(9.8)}`

t = √4.551

t = 2.13 s