Question

A lift is moveing with a uniform downward acceleration of $2m/s^2$ . A ball is dropped from height 2 cm from the floor of lift. Find the time take after which ball will strike the floor? $(g=10m/s^2)$

• A. $\sqrt{2}sec$
• B. 2 sec
• C. $\frac{1}{2}sec$
• D. $\frac{1}{\sqrt{2}}sec$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}sec$

Given that,

Acceleration of lift $a=2m/s^2$

We know that,

Acceleration of ball relative to lift = acceleration of ball relative to ground – acceleration of lift with respect to ground

$a^{\prime }=-10\hat{j}-(-2\hat{j})$

$a^{\prime }=8\hat{j}$

Now, from equation of motion

$s=ut-\frac{1}{2}a{t}^2$

$s=0+\frac{1}{2}\times 8\times t^2$

$t^2=\frac{1}{2}$

$t=\frac{1}{\sqrt{2}}s$

Hence, the time take after which ball will strike the floor is $\frac{1}{\sqrt{2}}s$