A metal plate weighing 200g is balanced in mid air by throwing 40 balls per second vertically upward from below.After collision ball rebounds with the same speed.What is the speed with which balls strike the plate.(mass of each ball-200g)?

0.122 meters per second

The plate's acceleration is ##9.8 m/(s^2)##. So, the balls must increase the plate's speed by ##9.8m/s##, so the plate will be balanced.

That means that each ball must increase the plate's speed in ##(9.8)/(40)=0.245 m/s##. As the total momentum is conserved in a rebound, if balls have a velocity ##v## the conservation of momentum is expressed as follows: ##0.2*v## (when the balls go up) ##= 0.2*0.245+(-v)*0.2## (the momentum of the plate going up plus the momentum of the ball going down).

Then ##v=0.2*0.245/0.4 = 0.122##.