First slide

Projection Under uniform Acceleration

Question

Time taken by the projectile to reach from I 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,
$u_{H} = u \cos 60^{0} = \frac{u}{2}$
AC = $u_{H} \times t = \frac{ut}{2}$
and AB = AC sec $30^{0}$
= $(\frac{ut}{2}) (\frac{2}{\sqrt{3}}) = \frac{ut}{\sqrt{3}}$

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