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Question

A ball is dropped from a height of 30 m in downward direction on the stationary floor. If the value of coefficient of restitution is 0.6 and ball rebounds back up to the height (h), find the height (h) after first collision.
(Take g=10 m/s2)

A
21.6 m
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B
22.8 m
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C
10.8 m
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D
26.8 m
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Solution

The correct option is C 10.8 m
The speed of ball at which it strike to the floor (u1)
u1=2gh=2×10×30=600 m/s
If u2 & v2 is the initial and final velocity of floor and since floor remains stationary,
u2=0 m/s and v2=0 m/s
Let v1 is the final velocity of ball after collision.
Coefficient of restitution,
e=velocity of separationvelocity of approach =v2v1u1u2
0.6=0(v1)6000
[ take velocity negative in +y direction]
v1=0.6×600 m/s
From energy conservation,
12mv21=mgh
h=v212g=(0.6×600)22×10 =0.6×0.6×6002×10
h=10.8 m

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