Question

# A parachutist drops first freely from an airplane for 10 s and then the parachute opens out. Now he descends with  net retardation of 2.5 m/s2. If he bails out of the plane at a height of 2495 m and g=10 m/s2, his velocity on reaching the ground will be5 m/s10 m/s15 m/s20 m/s

Solution

## The correct option is A 5 m/sThe velocity v acquired by the parachutist after 10 s, v=u+gt (By first equation of motion) =0+10×10  =100 m/s Distance covered by the parachutist during 10 s of falling freely is given by s1=ut+12gt2 (By second equation of motion) s1=0+12×10×102=500 m The distance travelled by the parachutist under retardation, s2=2495−500=1995 m Let vg be this velocity on reaching the ground. Then, v2g−v2=2as2 (By third equation of motion) v2g−(100)2=2×(−2.5)×1995 v2g=−(5×1995)+1002=25  vg=5 m/s

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