Given :
A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 174 meters.
To Find :
How long does it take for the ping pong ball to hit the ground ?
Solution :
We know, equation of motion of object doing free fall is :
[tex]s = ut + \dfrac{gt^2}{2}[/tex] ....1)
Here, g = 9.8 m/s² ( acceleration due to gravity )
Also, u = 0 m/s ( its free fall )
Putting these values in equation 1), we get :
[tex]174 = 0 + \dfrac{9.8\times t^2}{2}\\\\t = \sqrt{\dfrac{174\times 2}{9.8} }\ s\\\\t = 5.96 \ s[/tex]
Therefore, ball will take 5.96 s to hit the ground.
