"A 60 g tennis ball is dropped from a 2 m height, strikes a horizontal sidewalk and rebounds to a height of 1 m. Find the average force with which the ball hits the ground if the collision lasts 0.02s."

F = Changing momentum / time

Explanation:

change in momentum = final momentum - initial momentum

but before finding the force we have to find the initial and final vertical velocity of the ball in both cases (1st bounce and 2nd bounce) so first let's find the vertical velocities of the ball in the first bouncehere the ball dropped - it means (initial velocity is zero 0ms-1)but there is a final vertical velocity...let's find the final vertical velocity v² = u² + 2as (here a = g = 10ms^-2)v = √0²+2(10)2v = 6.32ms^-1After that let's find the vertical velocities of the ball in the second bounceso here our final velocity is zero (because vertical velocity in maximum heigh )is zeothen let's find our initial vertical velocityv² = u² + 2as0 = u² + 2(-10)1 (here -10 cause gravitational acceleration act positive only in the downward direction )u = √20u = 4.47ms^-1ok, now we found all the velocities then let's find the forceF = Changing momentum / timechanging momentum = impulse= mv-m(-u)= 0.06 (6.32+4.47)=0.6474 Nsso we found the chang in momentumthe let's find the forceF = Changing momentum / timeF = 0.6474 / 0.02F = 32.37N

Ihope you find the answer