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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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An aeroplane is moving with horizontal velocity u at a height h. The velocity of a packet dropped from it on the earth's surface will be

\[(a)\;\sqrt {u^2+2gh} \quad (b)\;\sqrt {2gh} \quad (c)\;2gh \quad(d)\;\sqrt {u^2-2gh} \]

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Initially the pocket has only horizontal velocity equal to u
When it reaches the ground it has horizontal velocity u and vertical velocity $\sqrt{2gh}$
Therefore resultant velocity $=\sqrt {u^2+2gh}$
Hence a is the correct answer


answered Jul 3, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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