Question

An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms.
the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is
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Answer 1

Answer:

A::B::C

Solution :

(a) If we consider elevator at rest, then relative acceleration of the bolt is

ar=9.8+1.2

=11m/s2 (downwards)

After 2 s, velocity of lift is v=at=(1.2)(2)=2.4m/s. Therefore, initial velocity of the bolt is

also 2.4m/s and it gets accelerated with relative acceleration 11m/s2. With respect to

elevator initial velocity of bolt is zero and it has to travel 2.7 m with 11m/s2. Thus, time

taken can be directly given as

2sa−−−√=2×2.711−−−−−−√

=0.7s.

(b) Displacement of bolt relative to ground in 0.7s.

s=ut+12at2

or s=(2.4)(0.7)+12(−9.8)(0.7)2

s=−0.72m

Velocity of bolt will become zero after a time

t0=ug

=2.49.8=0.245s

Therefore, distance travelled by the bolt=s1+s2=u22g+12g(t−t0)2

=(2.4)22×9.8+12×9.8(0.7−0.245)2

=1.3m