First slide

Projection Under uniform Acceleration

Question

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

2ut

Solution

Horizontal component of velocity,
$\begin{array}{l} u_{H} = ucos 60^{\circ} = \frac{u}{2} \\ ∴ AC = u_{H} t = \frac{ut}{2} \end{array}$
and AB = AC sec 30°
$= (\frac{ut}{2}) (\frac{2}{\sqrt{3}}) = ut / \sqrt{3}$

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