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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 be
  1. 5 m/s
  2. 10 m/s
  3. 15 m/s
  4. 20 m/s


Solution

The correct option is A 5 m/s
The 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=2495500=1995 m
Let vg be this velocity on reaching the ground.
Then, v2gv2=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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