Question

A ball is thrown horizontally from the top of a building 18.4 m high. The ball strikes the ground at a point 79.2 m from the base of the building. The acceleration of gravity is 9.8 m/s. Find the time the ball is in motion. Answer in units of s.

Answer 1

1.94 s

Step-by-step explanation:

To find the time of flight of the ball, we can just analyze its vertical motion.

The motion of the ball is a free fall motion, which is a uniform accelerated motion. Therefore we can use the suvat equation:

`s=ut+(1)/(2)at^2`

where

s is the vertical displacement

u = 0 is the initial vertical velocity of the ball

t is the time

`a=g` is the acceleration of gravity

Chosing downward as positive direction,

s = 18.4 m

`g=9.8 m/s^2`

And solving for t, we find the time during which the ball is in motion:

`t=\sqrt{(2s)/(g)}=\sqrt{(2(18.4))/(9.8)}=1.94 s`