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Q.

In figure shown below, the time taken by the projectile to reach from A to B is t and velocity of projection is u, then, the distance AB is equal to

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a

$\frac{\sqrt{3} u t}{2}$

b

$2 u t$

c

$\frac{u t}{\sqrt{3}}$

d

$\sqrt{3} u t$

answer is A.

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Detailed Solution

$u_{x} = u \cos θ = u \cos 60 °$

$x = A C = u_{x} . t; from figure \cos 30 ° = \frac{A C}{A B}$

$A B = \frac{A C}{\cos 30 °}$

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