Find the displacement covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.
A
0.7
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A0.7 Relative acceleration between the elevator car and the bolt is ar=ae+ab=ae+g Relative initial velocity between the elevator car and bolt is ur=u−u=0 where u is the velocity of the lift when the bolt drops from the elevator car. The bolt has also the same velocity u as that of the elevator car when it drops from the ceiling. Applying the 2nd kinematics equation, s=ut+12at2 S=urt+12art2=0+12art2=12×(1.2+9.8)×t2 ∴2.7=12×11×t2 ⇒t=√12=0.7 At the moment the bolt loses contact with the elevator, it has acquired an upward speed of 1.2×2=2.4m/s In the reference frame attached to the elevator shaft ( ground), the distance travelled by the bolt is 2.4×0.7−12×9.8×(0.7)2=−0.7m Hence, distance travelled by the bolt is 0.7m