Question

Do the algebraic solution first before plugging in the values.

2. An elevator ascends with an upward acceleration of 4.0 ft/s². At the instant its upward speed is 8.0 ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor. Calculate:
a. the time of flight of the bolt from ceiling to floor.
b. the distance it has fallen relative to the elevator shaft.

Answer 1

Downward direction as positive,

Its initial upward velociy u = -8ft/s.
Displacement = 9ft.
Acceleration due to gravity g = 32ft/s^2.

KINEMATICS EQUATION

v^2 = u^2 + 2as
v^2 = [-8] ^2 + 2*32*9
v = 25.3 ft/s

Time = displacement / average velocity = 9*2 / [25.3 - 8]
=1.04s.
Answer for A is 1.04s

PART B

It moves up with a velociy -8ft/s
Distance it moves up till its velocity becomes zero is
s = v^2 / 2g = 64/ 2*32 = 1 ft.

It falls through 1ft and then 9 ft.
Hence the distance it has fallen down is 9 +1 = 10 ft.