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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

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

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

edited Jan 26, 2014 by meena.p

+1 vote

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