A body is thrown up in a lift with a velocity $u$ relative to the lift and the time of flight found to be $t$ . The acceleration with which the lift is moving up is (Given acceleration due to gravity $= g$ )

\frac{u - g t}{t}

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\frac{2 u - g t}{t}

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\frac{u + g t}{t}

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\frac{2 u + g t}{t}

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Solution

The correct option is B $\frac{2 u - g t}{t}$
velocity of lift $v_{r e l} = u$
acceleration due to gravity $a_{g} = - g$
relative acceleration $a_{r e l} = - g - a$
$t_{r e l} = \frac{2 u}{g + a}$
$a = \frac{2 u - g t}{t}$
hence,option A is correct.

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Similar questions

Q. A body is thrown up in a lift with a velocity u relative to the lift and the time of flight is t. Show that the lift is moving up with an acceleration

\frac{2 u - t g}{t} .

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A body of mass m is thrown at an angle θ to the horizontal with the initial velocity u. Find the instantaneous power of gravity as function of time.

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A body is thrown up in a lift with a velocity u relative to the lift and the time of flight found to be t. The acceleration with which the lift is moving up is (Given acceleration due to gravity =g)

The correct option is B 2u−gttvelocity of lift vrel=uacceleration due to gravity ag=−grelative acceleration arel=−g−atrel=2ug+aa=2u−gtthence,option A is correct.

A body is thrown up in a lift with a velocity $u$ relative to the lift and the time of flight found to be $t$ . The acceleration with which the lift is moving up is (Given acceleration due to gravity $= g$ )

The correct option is B $\frac{2 u - g t}{t}$
velocity of lift $v_{r e l} = u$
acceleration due to gravity $a_{g} = - g$
relative acceleration $a_{r e l} = - g - a$
$t_{r e l} = \frac{2 u}{g + a}$
$a = \frac{2 u - g t}{t}$
hence,option A is correct.