Question

A 52kg ball is dropped from a height of 2,118mm. Calculate the height (above the ground) at which KE is one third of PE. Answer should be in standard unit.

Answer 1

let P be the point above the ground , where KE is equal to PE/3

   1/2 m v² = ( 1/3) m g H

         v = sqrt ( 2 ( 9.8 m/s² ) (2.118 m ) / 3 )

             =3.719 m/s

Velocity of ball at the ground is

      u = sqrt ( 2 g H )

          = sqrt ( 2 ( 9.8 m/s² ) ( 2.118 m ) )

           =6.443 m/s

Height above the ground at which KE is equal to PE /3

      u ² - v ² = 2 g h

           h = ( 6.443 m/s )² - (3.719 m/s)² / 2 (9.8 m/s² )

               = 1.41 m