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Question

A helicopter is rising up from ground with an acceleration of g m/s2, starting from rest after raising a height h it attains a velocity of v m/s. At this instant, a particle is now released from the helicopter. Take t=0 at releasing time, calculate the time t when particle reaches to the ground.

A
2hg
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B
22hg
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C
(1+2)2hg
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D
42hg
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Solution

The correct option is C (1+2)2hg
For upward motion of helicopter:

From third equation of motion,

v2=u2+2as

Here, u=0 ; v=v ; a=g ; s=h

v2=0+2as

v=2gh

Now, for the particle, motion will be as shown in figure, and it starts moving under gravity.


So, from second equation of motion,

s1=u1t+12a1t2

Here, u1=v=2gh ; s1=h ; a1=g

h=2ght12gt2

12gt22ghth=0

So, value of t will be,

t=2gh±2gh(4×g2×(h))2×g2

t=2hg(1±2)

Since, t can't be negative,

t=2hg(1+2)

Hence, option (C) is correct.

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