A lift is coming down with a downward acceleration of 2 $m / s^{2}$ . A boy standing in the lift thrown a stone vertically upwards with speed 5 m/s w.r.t. himself, the time after which he will catch the stone is
{ $g = 10 m / s^{2}$ }

2.5 s

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1.25 s

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3.25 s

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0.25 s

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Solution

The correct option is B 1.25 s
Given,

Downward acceleration $a = - 2 m / s^{2}$

Upward velocity $u = 5 m / s^{2}$

Relative acceleration of stone in frame of lift $a_{r} = g - a = - 10 - (- 2) = - 8 m / s^{2}$ (downward)

In frame of lift, magnitude of velocity of stone (when he catches it) is equal to the magnitude of projected velocity, $v = - u$ (equal magnitude opposite direction)

Apply kinematic equation

$v = u + a_{r} t$

$- u = u + a_{r} t$

$t = \frac{2 u}{a_{r}} = \frac{2 \times 5}{8} = 1.25 sec$

Time after which he catches the stone is $1.25 sec$

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A lift is coming down with a downward acceleration of 2 m/s2. A boy standing in the lift thrown a stone vertically upwards with speed 5 m/s w.r.t. himself, the time after which he will catch the stone is {g=10 m/s2}

A lift is coming down with a downward acceleration of 2 $m / s^{2}$ . A boy standing in the lift thrown a stone vertically upwards with speed 5 m/s w.r.t. himself, the time after which he will catch the stone is
{ $g = 10 m / s^{2}$ }