A 70-g ball B dropped from a height h_0=1.5 m reaches a height h_2=0.25 m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine

Question

(a) the coefficient of restitution between the ball and the plates,
(b) the height h_1 of the ball’s first bounce.

Answer 1

I do not care what the mass is, call it m
If it bounces off the same thing twice (I do not get it about the two plates because you provide data for only one of them)
v = velocity at impact with plate A
(1/2) m v^2 = m g h
v = sqrt(2 gh) = sqrt (2*9.81*1.5) = sqrt (29.43) = 5.42 m/s

w = velocity headed up from first hit = 5.42 k
where k is our coef of restitution

we lose no energy going up and then coming down so
we hit plate B at speed w = 5.42 k
we rebound from plate B at speed x = 5.42 k^2

now
m g h = m g (0.25) = (1/2) m (5.42k^2)^2
9.81 (.25) = .5 (29.4) k^4
k^4 = .167
k^2 = .408
k = .639
That should get you started although I am not sure what the two plates is about.