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Time taken to reach A to B for a projectile is 't'. Then the distance AB is equal to


$ a)\;\large\frac{ut}{\sqrt 3} \\ b)\;\large\frac{\sqrt 3 ut}{2}\\ c)\;\sqrt ut \\ d)\;2 ut$

1 Answer

Horizontal component of u is $u \cos 60$
$AB=AC \sec 30$
$\qquad=\large\frac{ut}{2} \frac{2}{\sqrt 3}$
$\qquad=\large\frac{ut}{\sqrt 3}$
Hence a is the correct answer.


answered Jun 26, 2013 by meena.p
edited Jan 25, 2014 by meena.p

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