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

Answer: (A)

Horizontal component of velocity
$ \text{u}_{\text{H}} = \text{u cos 60}^{\text{o}} = \frac{\text{u}}{2}$
$\therefore \quad \mathrm{AC}=\left(\mathrm{u}_{\mathrm{H}}\right) \mathrm{t}=\frac{\mathrm{ut}}{2}$

and $AB \, = \, AC \, sec \, 30^\circ $
$ = \left(\frac{\text{ut}}{2}\right) \left(\frac{2}{\sqrt{3}}\right) = \frac{\text{ut}}{\sqrt{3}}$