A paruchutist drops first freely from an aeroplane for 10 s and then his parachute opens out. Now he descends with a net retardation of 2.5 ms-2. If he bails out of the plane at a height of 2495 m and g = 10 ms-2, his velocity on reaching the ground will be

The velocity v acquired by the parachutist after 10 s:v=u+gt=0+10×10=100ms−1 Then, s1=ut+12gt2=0+12×10×102=500mThe distance travelled by the parachutist under retardation iss2=2495−500=1995mLet vg be the velocity on reaching the ground. Thenvg2−v2=2as2 or  vg2−(100)2=2×(−2.5)×1995 or vg=5ms−1