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Question

A person throws vertically 10 balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

A
10 m
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B
5 m
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C
5 cm
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D
10 cm
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Solution

The correct option is C 5 cm
10 balls are thrown per second.
Time interval between two balls thrown =110 s
This is the time that a ball takes to reach its highest point.
Using first equation of motion
v=u+at
As the velocity of ball is 0 at its maximum height
0=ug(110) (By first equation of motion)
u=g10
Height achieved by the balls is given by
v2u2=2aS
02(g10)2=2×(g)h
h=g2×102=0.05 m=5 cm
The answer is (c)

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