A parachutist 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 = 100 ms-1Then,  s1=ut+12gt2=0+12 × 10 × 102 = 500mThe distance travelled by the parachutist under retardation iss2= 2495 - 500 = 1995 m Let vg be the velocity on reaching the ground. Thenvg2 -v2= 2as2 or vg2-(100)2 = 2 × (-2.5) × 1995  or vg=5 ms-1