An elevator is goning vertically up 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 $60^{o}$ . The time taken by the ball to return the floor is $(g = 10 m / s^{2})$ .

\frac{1}{3} sec

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1 sec

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\frac{2}{\sqrt{3}} sec

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\sqrt{3} sec

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Solution

The correct option is A $\frac{1}{3} sec$
Components of velocity of ball relative to lift are $:$
$u_{x} = 4 cos 30^{0} = 2 \sqrt{3} m / s$
$u_{y} = 4 sin 30^{0} = 2 m / s$
$T = \frac{2 u_{y}}{12} = \frac{u_{y}}{6} = \frac{2}{6} = \frac{1}{3} s$
Hence,
option $(A)$ is correct answer.

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A very broad 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 lift with a speed of $4 m / s$ relative to the floor at an elevation of $30^{\circ}$ . The time taken by the ball to return the floor is ( $g$ = $10 m / s^{2}$ ) from time of projection.