An aeroplane is moving with constant horizontal velocity u at height h. The velocity of a packet dropped from aeroplane, when it reaches on the earth's surface will be (g is acceleration due to gravity)

\sqrt{u^{2} + 2 g h}

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\sqrt{2 g h}

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2 g h

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\sqrt{u^{2} - 2 g h}

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Solution

The correct option is A $\sqrt{u^{2} + 2 g h}$
Here, since there are no external forces and no non conservative forces, we can conserve the total mechanical energy of the system in the initial and final cases.
Assuming the mass of the packet to be $m$ ,
For when the aeroplane is carrying the packet.
$T M E_{i} = K E_{i} + P E_{i}$
$T M E_{i} = \frac{1}{2} m u^{2} + m g h$

When the packet reaches the ground, all the potential energy is converted into its kinetic energy. ( $P E_{f} = 0$ )
$T M E_{f} = K E_{f} + P E_{f}$
$T M E_{f} = \frac{1}{2} m v^{2}$

SInce, $T M E_{i} = T M E_{f}$
$\frac{1}{2} m v^{2} = \frac{1}{2} m u^{2} + m g h$

$\frac{1}{2} v^{2} = \frac{1}{2} u^{2} + g h$

$v^{2} = u^{2} + 2 g h$

$v = \sqrt{u^{2} + 2 g h}$

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