Question

An open lift is moving upwards with velocity $10m/s$. It has an upward acceleration of $2m/s^2$. A ball is projected upwards with upwards with velocity $20m/s$ relative to the ground. Find the time when the ball again meets the lift.

• A. $\frac{5}{3}s$
• B. $\frac{4}{3}s$
• C. $\frac{3}{4}s$
• D. $\frac{1}{4}s$

Answer 1

The correct option is A $\frac{5}{3}s$

Let the time of Height of the ball be $t$. Then. with respect to the lift.

$V_{B/L}=V_B-V_L=20-10=10m/s$

$a_{B/L}=a_B-a_L=(-10)-\left(2\right)=-12m/s^2$

To find $t$, applying second $e{q}^n$ to motion

$h=u\left(t\right)+a{t}^2$

$\Rightarrow o=\left(10\right)\left(t\right)+\frac{(-12)t^2}{2}$

$\Rightarrow b{t}^2=10t$

$\Rightarrow t=5/3^3$

Option $-A$ is correct.