A parachutist drops freely from an airplane for 10 s before the parachute opens. He then descends with a uniform retardation of 2.5 ms-2. If he bails out of the plane at a height of 2495 m and g is 10 ms-2, his velocity on reaching the ground will be

Initial velocity of dropping = zero Let v1 be velocity at end of 10 s⇒v1=gt⇒v1=100ms−1Distance travelled during this time ish1=v122g=(100)22(10)⇒h1=500mSo, a remaining distance of 2495-500 = 1995 m has to be travelled with a retardation of 2.5 ms-2. Let the parachutist strike the ground with velocity v.Then v2−v12=2ah−h′⇒v2−(100)2=2(−2.5)(1995)⇒v2=10000−9975⇒v2=25 ⇒v=5ms−1