A parachutist of weight w strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3g. The force exerted on him by the ground during landing is:
A
w
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B
2w
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C
3w
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D
4w
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Solution
The correct option is D4w The parachutist goes upward with an acceleration 3g and his weight is w. Therefore, w=mg N−mg=m(3g)⇒N=4mg=4w.