Question

An elevator is going up vertically with a constant acceleration of 2 m/s${}^2$. at the instant when its velocity is 4 m/s a ball is projected from the floor of the elevator with a speed of 4 m/s relative to the floor with an angular elevation of ${30}^{\circ }$. the time taken by te ball to return to the floor is :(take g=10$m/s^2$)

• A. $\frac{1}{3}\sec$
• B. $\frac{1}{\sqrt{3}}\sec$
• C. $\frac{2}{\sqrt{3}}\sec$
• D. $\sqrt{3}\sec$

Answer 1

The correct option is A $\frac{1}{3}\sec$

Given

Acceleration of Elevator $=2m/s^2$

Initial velocity of elevator $=4m/s$

Velocity of ball $=4m/s$

Angle of projection $={30}^{\circ }$, $g=10m/s^2$

To Find

Time taken to return the ball to elevator = ?

Ref. image

Geff $=g-(-a_e)$

$=10-(-2)$

$=12m/s^2$

$s_y=0\to$ displacement in y direction

$u_y=u\cos {60}^{\circ }=4\times \frac{1}{2}=2m/s$

$S_y=u_{y}t+\frac{1}{2}geff{t}^2$

$0=2\times t-\frac{1}{2}\times 12\times t^2$

$t=\frac{1}{3}sec$