A ball of mass 1kg is dropped from height 9.8m strikes with ground and rebound at height of 4.9m if the time of contact b/w ball and ground is 0.1 sec then find impulse and average force acting on the ball?

Velocity of the ball when it hit the ground can be calculated as 12mv2=mgh⇒v=2gh−−−√=2x10x20−−−−−−−√⇒v=20m/s in downward direction.on striking the ground let the ball rebound with the velocity u in upward direction. it can be calculated as 12mu2=mgh⇒u=2gh−−−√=2x10x5−−−−−−√⇒u=10m/s in upward directionChange in momentum of the ball = final momentum of the ball −initial momentum of the ball = mu⇀−mv⇀⇒m( u +v) as the direction of u and v are opposite∆P=1(10+20) = 30kgm/s