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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
askIITians Faculty 181 Points
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