Question

An elevator (lift) ascends with an upward acceleration of 1.2ms^-2. At the instant when its upward speed is 2.4 ms^-1, a loose bolt drops from the ceiling of the elevator 2.7m above the floor of the elevator. Calculate (a) the time of flight of the bolt from the ceiling to the floor and (b) the distance it has fallen relaative to the elevator shaft.

Answer 1

Answer:

$t = \sqrt{2h \div g}$

here g=g+a because it is moving upward

so t=√2h÷(g+a)

so t=0.7

the given instant velocity

so s=-ut+1/2at²

-ve indicate it s downward

s=-2.4×0.7+1/2×10×0.7²

=2.48-1.68

=0.7m