A ball after freely falling from a height of 4.9m strikes a fixed horizontal plane. If the coefficient of restitution is 34, after what time interval the ball will strike the plane second time :
A
0.5s
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B
1s
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C
1.5s
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D
2s
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Solution
The correct option is C1.5s According to question,
Velocity of ball just before striking plane, v
using, v2−u2=2as ⇒v=√2×9.8×4.9=9.8m/s ...(1)
Since horizontal plane remains fixed, initial and final velocity of the plane remains same.
Rebound velocity, v′=ev =34×v [e=3/4, given] =34×9.8 [from (1)] =7.35m/s..... (2)
Let us suppose, after ′t′ time ball again strikes the plane
As ball reaches to same position, its displacement is zero.
Using, s=ut+12at2 ⇒0=7.35t−12×9.8×t2 [from (2), u=v′] ⇒t=7.35×29.8 ⇒t=32=1.5s
After 1.5s ball will strike the plane for the second time.