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

Answer:

t = 2.13 s

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

$s = u t + \dfrac{1}{2}gt^2$

initial velocity is zero

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 22.3}{9.8}}$

       t = √4.551

        t = 2.13 s