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. $2ut$

Answer 1

The correct option is A $\frac{ut}{\sqrt{3}}$


Angle of projection with the horizontal $={60}^{\circ }$
Horizontal distance covered in time $t$
$=u\cos {60}^{\circ }\times t=AB\cos {30}^{\circ }$ (by geometry)
$\therefore AB=\frac{ut}{\sqrt{3}}$