An aeroplane drops a parachutist. After covering a distance of 40 m, he opens the parachute and retards at 2 ms -2 . If he reaches the ground with a speed of 2 m -1 , he remains in the air for about
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Verified by ExpertsThe correct answer is:
A
Using h = 1/2 gt 2 , we get, t 1 = 
.
Let t 1 be the time taken from instants of jumping to the opening of parachute, then
t 1 = 
= 2.86 s
His velocity at this point is given by,
v 1 2 = 2gh 1 = 2 × 9.8 × 40
784 or v 1 = 28 ms -1
For the remaining journey ,
v = v 1 + at 2
or t 2 = 
∴ ∴ total time = t 1 + t 2 = 2.86 + 13
= 15.86 = 16s
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