Q.
The time taken by the projectile to reach from $A$ to $B$ is $t,$ then the distance $AB$ is equal to
A) $2\ ut$
B) $\sqrt{3}\ ut$
C) $\frac{\sqrt{3}}{2}ut$
D) $\frac{ut}{\sqrt{3}}$
Explanation
(d) Horizontal component of velocity at $A$
[image:4_image116.emf]
$v_{H} = u\cos{60{^\circ}} = \frac{u}{2}\ \therefore AC = u_{H} \times t = \frac{ut}{2}$
$AB = AC\sec{30{^\circ}} = \frac{ut}{2} \times \frac{2}{\sqrt{3}} = \frac{ut}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.