Question

A lift moves upward with an acceleration of 1.2 ms^-2. A nail falls from the ceiling of the lift 3 m above the floor of the lift, Distance of its fall with reference to the shaft of the lift is

• A. 0.75 m
• B. 0.5 m
• C. 1 m
• D. 1.5m

Answer 1

Answer: (C)

1 m

Answer 2

Let y₁be the distance covered by the fan and y₂ be the distance covered by lift just before the fan reaches the floor of the lift.

∴ y₁ + y₂ = 3

but y₁ = 2.4t + 1/2 9.8 t²

(initial velocity of fan is upward and acceleration is downward)

and y₂ = 2.4t + 1/2 × 1.2 t²

= 2.4t + 0.6t²

∴ y₁ + y₂ = 4.9t² + 0.6t² = 5.5t²

or 3 = 5.5 t²

or t = $\sqrt { 3 / 5.5 }$ = 0.74 s

Now y₁ = -2.4 × 0.74 + 4.9 × 0.74²

= -1.68 + 2.68 = 1 m