If a ball of mass 0.1 kg hits the ground from the height of 20 m and bounce back to the same height, then find out the force exerted on the ball if the time of impact is 0.04 sec . (Take, g=10 ~m / s ^2 )

Step-by-step solution

Here, mass of ball m=0.1 ~kg Velocity attained by the ball before hitting the aligned ground ( v ) & = 2 gh (- j ) & = 2 10 20 (- j )=-20 j / s aligned Velocity of ball when bounce back to the same height after hitting the ground, V^ =-v=-(-20 j )=20 j ~m / s Change in velocity V=V^ -V=20 j -(-20 j )=40 j ~m / s Force exerted on the ball aligned f & = P t = m V 0.04 [ given t =0.04] & = 0.1 40 j 0.04 =100 ~N ( j ) aligned