Question

A lift moves upward with an acceleration of $1.2m{s}^{-2}$. Two seconds after the lift starts, a nail falls from the ceiling of the lift $3m$ above the floor of the lift. Distance of its fall with reference to the shaft of the lift is

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

Answer 1

The correct option is C $1m$

Let the distance covered by the nail and the lift be $s_1$ and $s_2$.
Thus we get
$s_1+s_2=3$
Now distance travelled by the nail is given as
$s_1=-2.4t+\frac{1}{2}\times 9.8t^2$
Distance travelled by the lift in the same time is given as
$s_2=2.4t+\frac{1}{2}\times 1.2t^2$
Thus we get
$s_1+s_2=(0.6+4.9)t^2=5.5t^2$
or
$t=\sqrt{\frac{3}{5.5}}=0.74s$
Thus we get
$s_1=-2.4t+\frac{1}{2}\times 9.8t^2$
or
$s_1\approx 1m$