Question

A body is projected with a velocity $u$ at an angle of ${60}^o$ to the horizontal. The time after which it will be moving in a direction of ${30}^o$ to the horizontal is

• A. $\frac{1}{\sqrt{3}}\frac{u}{g}$
• B. $\frac{\sqrt{3}u}{g}$
• C. $\frac{\sqrt{3}u}{2g}$
• D. $\frac{2u}{\sqrt{3}g}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}\frac{u}{g}$

$u\left(t\right)=u_x\hat{i}+u_y\hat{j}$
$u\left(t\right)=\left(u\cos \theta \right)\hat{i}+(u\sin \theta -gt)\hat{j}(\theta ={60}^0)$
Now, for this to make ${30}^0$ with horizontal.
$\tan {30}^0=\frac{u_y}{u_x}=\frac{u\sin {60}^0-gt}{u\cos {60}^0}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\frac{\sqrt{3}}{2}u-gt}{\frac{4}{2}}\Rightarrow u(\frac{\sqrt{3}}{2}-\frac{1}{2\sqrt{3}})=gt$

$\Rightarrow \left|t\right|=\left|\frac{u}{g}\right(+\frac{1}{\sqrt{3}}\left)\right|$

$\Rightarrow t=\frac{u}{g}\frac{1}{\sqrt{3}}$