A life of very broad-floor is moving vertically upward with a constant retardation equal to ' $g$ '. At an instant a stone is projected from a point on the floor of the lift at at angle of elevation. Then the trajectory of the stone is

A Parabola in the life frame

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A straight line in the lift frame

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A Parabola in the ground frame

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A Straight line in the ground frame

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Solution

The correct options are
A A straight line in the lift frame
B A Parabola in the ground frame
The stone will move in straight line with respect to the lift because same acceleration g is acting on both the bodies

we have $s = u t + \frac{1}{2} g t^{2}$

This is a parabolic equation, so with respect to the ground the ball will move in parabolic path.

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Similar questions

Q. 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^{o}

. The time taken by the ball to return the floor is
(

g = 10 m / s^{2}

)

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.