First slide

Projection Under uniform Acceleration

Question

In Fig., 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

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

$\sqrt{3} ut$
D

2ut

Solution

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

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