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Question
A packet is dropped from a stationary helicopter, hovering at a height ‘h’ from ground level, reaches the ground in 12s. Calculate
- the value of h
- final velocity of packet on reaching the ground. (Take g = 9.8 ms−2)
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Solution
Height of the helicopter = h =?
Initial velocity = u = 0
Time = t = 12s
Acceleration = a = + g = + 9.8 ms−2
(1) S = ut + `1/2` at2
h = `0(12)+1/2(9.8)(12)^2`
h = 0 + 4.9 (144)
h = 705.6 m
(2) Let v = velocity of the packet on reaching the ground.
v = u + at
v = 0 + (9.8) 12
v = 117.6 ms−1
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