A parachutist drops freely from an aeroplane for 10s before the parachute opens out. then he descend

Question

A parachutist drops freely from an aeroplane for 10s before the parachute opens out. then he descends with a net retardation of 2.5 m/s2. If he bails out of plane at at of 2495 m and g = 10 m/s2, hit velocity on reaching the ground will be :-

Accepted Answer

Velocity after 10 sec is equal to
v = u+ at

v = 0 + (9.8) (10) = 100 m/s
Distance covered in 10 sec is equal to
^_s=(10) (10)² = 500 m
Now from v² = u² + 2as.
⇒ v² = (100)² - 2 (2.5) (2495 - 500) = 25 ⇒ v = 5 ms^-1

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