Question

A parachutist drops freely from an aeroplane for 10
seconds before the parachute opens out. Then he de-
scends with a net retardation of 2 m/sec. His velocity
when he reaches the ground is 8 m/sec. Find the height
at which he gets out of the aeroplane ?​

Answer 1

$\huge\pink{\overbrace{\purple{\underbrace{\blue{\mathbf{:Answer:}}}}}}$

• 2875 m

$\huge\pink{\overbrace{\purple{\underbrace{\blue{\mathbf{Explanation:}}}}}}$

Distance he falls before parachute opens is

S1 = (1/2) x g x 100 = 490 m

Then his velocity is

u= gt = 9.8 x 10 = 98.0 m/s

Velocity on reaching ground

===> v =8m/s

Retardation

===>2= m/s²

v²u² = 2aS2 ====> S2= 2385 m

Total distance;

S = S1+S2 = 2385 + 490

====> S= 2875 m .........

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