A lift is moving up with acceleration a- A person inside the lift throws the -160-ball upwards with a velocity u relative to hand- -160-What is the maximum height reached by the ball in the lift- u-2-2-g - a- u-2-g - a- u-2-g - a- u-g - a-

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A lift is mov...

Question

A lift is moving up with acceleration a. A person inside the lift throws the ball upwards with a velocity u relative to hand. What is the maximum height reached by the ball in the lift?

\frac{u^{2}}{2 (g + a)}

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\frac{u}{2 (g + a)}

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\frac{u^{2}}{(g + a)}

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\frac{u}{(g + a)}

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Solution

The correct option is A $\frac{u^{2}}{2 (g + a)}$
$v^{2} - u^{2} = 2 a s$
Relative acceleration $= a$
Relative initial velocity $= u$
At maximum height in the lift, ball is stationary when seen from lift
ie. Final velocity $= 0$
$0 - u^{2} = - 2 (g + a) H$
$H = \frac{u^{2}}{2 (g + a)}$

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A lift is moving up with acceleration a. A person inside the lift throws the ball upwards with a velocity u relative to hand. What is the maximum height reached by the ball in the lift?

The correct option is A u22(g+a)v2−u2=2asRelative acceleration =aRelative initial velocity =uAt maximum height in the lift, ball is stationary when seen from liftie. Final velocity=00−u2=−2(g+a)H H=u22(g+a)

The correct option is A $\frac{u^{2}}{2 (g + a)}$
$v^{2} - u^{2} = 2 a s$
Relative acceleration $= a$
Relative initial velocity $= u$
At maximum height in the lift, ball is stationary when seen from lift
ie. Final velocity $= 0$
$0 - u^{2} = - 2 (g + a) H$
$H = \frac{u^{2}}{2 (g + a)}$