A particle is projected with a velocity u making an angle $θ$ with the horizontal. At any instant, its velocity V is at right angles to its initial velocity u ; then V is

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$ucos θ$

$utan θ$

$ucot θ$

$u \sec θ$

answer is C.

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Detailed Solution

Because horizontal component of velocity remains constant in projectile motion, hence
$ucos θ = Vcos (90^{\circ} - θ)$
$ucos θ = Vsin θ or V = ucot θ$

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