Question

Time taken by the projectile to reach from A to B is t. Then the distance AB is equal to :

• A. $\frac{ut}{\sqrt{3}}$
• B. $\frac{\sqrt{3}ut}{2}$
• C. $\sqrt{3}$ut
• D. 2 ut

Answer 1

Answer: (A)

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

Answer 2

Horizontal component of velocity

u_H = u cos 60⁰ =$\frac{u}{2}$

\ AC = (u_H)t = $\frac{ut}{2}$

and AB = AC sec 30⁰ =$\left( \frac{ut}{2} \right)$=$\left( \frac{2}{\sqrt{3}} \right)$=$\frac{ut}{\sqrt{3}}$