A very broad elevator is going up vertically with a constant acceleration of 2ms- 2 . At the instant when its velocity is 4 ms- 1 a ball is projected from the floor of the elevator with a speed of 4 ms- 1 with respect to the elevator at an angle of 30° . The time taken by the ball to return the floor is (.g=10 ms- 2.)

Question

A very broad elevator is going up vertically with a constant acceleration of 2ms- 2 . At the instant when its velocity is 4 ms- 1 a ball is projected from the floor of the elevator with a speed of 4 ms- 1 with respect to the elevator at an angle of 30° . The time taken by the ball to return the floor is (.g=10 ms- 2.) (A) 1/2 s (B) 1/3 s (C) 1/4 s (D) 1 s

Tardigrade Admin · Accepted Answer

Components of velocity of ball relative to lift are
Solution

u_{x} = 4 cos 3 0^{\circ} = 23 m s^{- 1}

and

u_{y} = 4 s in 3 0^{\circ} = 2 m s^{- 1}

and acceleration of ball relative to lift is

12 m s^{- 2}

in negative

y

-direction or vertically downwards. Hence, time of flight

T = \frac{2 u _{y}}{12} = \frac{u _{y}}{6} = \frac{2}{6} = \frac{1}{3} s

1/2 s

1/3 s

1/4 s

1 s

Components of velocity of ball relative to lift are ux​=4cos30∘=23​ms−1 and uy​=4sin30∘=2ms−1 and acceleration of ball relative to lift is 12ms−2 in negative y -direction or vertically downwards. Hence, time of flight T=122uy​​=6uy​​=62​=31​s