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

and $AB=ACsec30^{\circ }$
$=\left(\frac{\text{ut}}{2}\right)\left(\frac{2}{\sqrt{3}}\right)=\frac{\text{ut}}{\sqrt{3}}$