Question : A 3 meter tall elevator moving at 2 m/s upward has a bolt come loose at t=0. The bolt “freefalls” for a period of time and finally hits the elevator’s floor. How far has the elevator’s floor moved during the time the bolt free fell?
Doubt by Dev
Solution :
Let the total time taken by the bolt to hit the elevator's floor be t.
For elevator
a = 0
u = v = 2 m/s
Distance travelled by the elevator in the upward direction is S1
Using S=ut+½at2
S1 = 2t + ½(0)t2
S1 = 2t — (1)
For Bolt
u = - 2/s
a = g = 9.8 m/s
t = ?
Distance travelled by the Bolt in the downward direction will be S2
S2 = (-2)t + ½(9.8)t2 — (2)
Also,
S1 + S2 = 3m
Substituting the values from eq (1) and (2)
2t + (-2t) + ½(9.8)t2 = 3
9.8 t2 = 6
t2 = 6/9.8
t2 = 60/98
t2 = 30/49
t2 = 0.612
t = 0.78 s
Putting in eq (1)
S1 = 2(0.78)
S1 = 1.56 m
Hence, the elevator’s floor moved up by a distance of 1.56 m during the time the bolt free fall.
Similar Question : An elevator ascends with an upward acceleration of 1.2 m/s2. At the instant when its upward speed is 2.4 m/s, a loose bolt drops from the ceiling of the elevator 2.7 m above the floor of the elevator. Calculate (a) time of flight of the bolt from the ceiling to the floor and (b) the distance it has fallen relative to the elevator shaft.
[(a) 0.7 s (b)0.72m]