First slide

Projectile motion

Question

In figure, the time taken by the projectile to reach from A to B is t. Then the distance AB is equal to

Moderate

A

$\frac{ut}{\sqrt{3}}$
B

$\frac{\sqrt{3} ut}{2}$
C

$\sqrt{3} ut$
D

$2 ut$

Solution

Horizontal component of velocity, $u_{H} = ucos 60^{\circ} = \frac{u}{2}$
$AC = u_{H} \times t = \frac{ut}{2}$
and $AB = ACsec 30^{\circ} = (\frac{ut}{2}) (\frac{2}{\sqrt{3}}) = \frac{ut}{\sqrt{3}}$

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