Question

A very broad elevator is going up Vertically with a constant acceleration of $2m/s^2$. At the instant when its velocity is $4m/s$ a ball is projected from the floor of the lift with a speed of $4m/s$ relative to the floor at an elevation of $30^{\circ }$. Time taken by the ball to return the floor is $(g=10m/s^2)$

• A. $\frac{1}{2}s$
• B. $\frac{1}{3}s$
• C. $\frac{1}{4}s$
• D. $1s$

Answer 1

Answer: (B)

Components of velocity of ball relative to lift are:

$u_x=4cos30^{\circ }=2\sqrt{3}m/s$

$u_y=4sin30^0=2m/s$

and acceleration of ball relative to lift is $12m/s^2$ in negative y-direction or vertically downwards. Hence time of flight

$T=\frac{2u_y}{12}=\frac{u_y}{6}=\frac{2}{6}=\frac{1}{3}s$