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# a lift accelerates downwards from rest at rate of 2m/s^ starting 100m above the ground after 3 sec an object falls out of the lift which will reach the ground first what is the time interval between their striking the ground

Apoorva Arora IIT Roorkee
6 years ago
here we use the three equations of motion, i.e.
$v=u+at$
$s=ut+\frac{1}{2}at^{2}$
$v^{2}-u^{2}=2as$
here after 3 sec the velocity of the lift becomes 6 m/s according to eq 1, therefore when we drop the object, its initial velocity u=6 m/s. also, acc to the eq 2, the lift has traveled 9 m. therefore, the rest distance left is 91 m.
for the object, u= 6 m/s, s= 91 m and a=g (take 10 m/s(square)) and hence from eq 2 t= 3.7 sec (approx)
similarly for the lift, u= 6 m/s, s= 91 m and a= 2 m/s(square) and hence t= 7 sec
therefore the object hits the ground first and then the lift after a difference of 3.3 sec

Thanks and Regards
Apoorva Arora
IIT Roorkee