Question

In a lift moving up with an acceleration of $5m{s}^{-2}$, a ball is dropped from a height of $1.25$m. The time taken by the ball to reach the floor of the lift is .....(nearly) $(g=10m{s}^{-2})$.

• A. $0.3$ second
• B. $0.2$ second
• C. $0.16$ second
• D. $0.4$ second

Answer 1

Answer: (D)

$0.4$ second

Given : $g=10$ $m/s^2$ $S=1.25$ m $a=5$ $m/s^2$

Effective acceleration due to gravity experienced by ball in lift frame, $g_{eff}=a+g$

$\therefore$ $g_{eff}=5+10=15$ $m/s^2$

Initial speed of ball $u=0$

Using $S=ut+\frac{1}{2}{g}_{eff}{t}^2$

$\therefore$ $1.25=0+\frac{1}{2}\times 15t^2$ $⟹t=0.4$ s