Question

a loose nut from a bolt on the bottom of an elevator which is moving up the shaft at 3m/s falls freely.the nut strike the bottom of the shaft in 2second.Distance of elevator from the bottom of the elevator from the bottom of shaft when the nut fell off is

Answer 1

u=3m/s

v=0m/s

t=2sec

s=?

a=v-u/t

=0-3/2= -1.5m/s^2

According to third equation of motion, v^2-u^2=2as

(0)^2-(3)^2=2X-1.5Xs

0-9= -3s

-9= -3s

3s=9

s=9/3=3m

Distance travelled=3m

Answer 2

ANSWER

s=ut−21gt2

H=3t−21(9.8)t2

t=2 s

H=3(2)−21(9.8)(2)2

⇒H=−13.6 m