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Question

A hot air balloon, released from ground, moves up with a constant acceleration of 5 ms2. A stone is released from it at the end of 8s. Find the height (in m) of the stone from the ground, 8s after it is released from the balloon. (Take g=10 ms2)

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Solution

The velocity of the stone, when it is released from the balloon is given by, v=u+at and is equal to the velocity of the balloon. Consider downward direction as position and substitute,
u=0, a=5 ms2, t=8
v=5×8=40 ms1
The displacement of the stone after 8s is
s=ut+12gt2
Substituting s=h, u=40 ms1
t=8s, g=10 ms2
h=40×8+12×10×82
h=320+320=0

The stone returns back to the position from where it was released, i.e, the height of the balloon from the ground at the end of 8s.
s=ut+12at2
substitute, u=0
a=5 ms2, t=8s
s=12×5×8×8=5×4×8
s=160 m
The height of the balloon from the ground is 160 m

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