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Standard XII

Physics

Frames of Reference

An open eleva...

Question

An open elevator is ascending with zero acceleration and speed $10 m / s .$ A ball is thrown vertically up by a boy (boy is in elevator) when he is at a height $10 m$ from the ground, the velocity of projection is $30 m / s$ with respect to elevator. Choose correct option(s) assuming height of the boy very small: $(g = 10 m / s^{2})$

Maximum height attained by the ball from ground is

90 m .

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Maximum height attained by the ball with respect to lift from the point of projection is

45 m .

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Time taken by the ball to meet the elevator again is

6

sec

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The speed of the ball when it comes back to the boy is

20 m / s

with respect to ground

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Solution

The correct option is A Maximum height attained by the ball from ground is $90 m .$
$u^{2} = u^{2} + 2 a s$
$\frac{2 u^{2}}{2 a} = s$
$s = \frac{30 \times 30}{10} = (90) m$

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The correct option is A Maximum height attained by the ball from ground is 90m.u2=u2+2as2u22a=ss=30×3010=(90)m

The correct option is A Maximum height attained by the ball from ground is $90 m .$
$u^{2} = u^{2} + 2 a s$
$\frac{2 u^{2}}{2 a} = s$
$s = \frac{30 \times 30}{10} = (90) m$