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Question

A balloon is released from the ground (h=0 m) from rest and starts accelerating with acceleration 5 m/s2 vertically upwards. An object is dropped from the balloon, when it is at height 10 m from the ground. If the balloon is released at t=0 sec, then find the time when the object reaches the ground. Take g=10 m/s2.

A
2.73 s
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B
4.73 s
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C
4 s
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D
2 s
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Solution

The correct option is B 4.73 s
When object is dropped from balloon, h=10 m
and let velocity of balloon be v m/s
Given a=+5 m/s2
Taking vertically upwards as +ve y-direction and applying kinematic equation:
v2=u2+2ah
v2=0+2×5×10
v2=100v=10 m/s

Using 2nd eqn of motion
h=ut1+12at21
+10=0+12×5×t21
t21=4
or t1=2 s

For released object, initial velocity will be equal to that of velocity of balloon
u=10 m/s
Let it reach the ground after time t2 s.
Then, displacement of object S=10 m
Applying 2nd eqn of motion
S=ut2+12at22
10=+10t212×10×t22 [a=g=10 m/s2]
5t2210t210=0
t2=1+3=2.73 s

Hence, total time t=t1+t2=2+2.73=4.73 s

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