A particle is projected upwards with a velocity of $100\text{m/sec}$ at an angle of ${60}^{\circ }$ with the vertical. Find the time when the particle will move perpendicular to its initial direction taking $g=10{\text{m/sec}}^2$.

The correct option is D $20\text{sec}$
Let say at point $\text{P}$ velocity of particle makes ${90}^{\circ }$ with initial velocity ($u=100\text{m/s}$ as shown in the figure below.


Let us consider, it takes $t$ $\text{sec}$ to reach the point $\text{P}$.


Applying the equation of motion along $\text{y-axis}$,

$-(v_p{)}_y=100\sin {30}^{\circ }-gt$

$\Rightarrow -(v_p{)}_y=50-10t$

$\Rightarrow (v_p{)}_y=(10t-50)$

And applying equation of motion along $\text{x-axis}$,

$(v_p{)}_x=100\cos {30}^{\circ }=50\sqrt{3}$

and from figure,

$\tan {60}^{\circ }=\frac{(v_p{)}_y}{(v_p{)}_x}=\frac{10t-50}{50\sqrt{3}}$

$\Rightarrow \sqrt{3}=\frac{10t-50}{50\sqrt{3}}$

$\Rightarrow 150=10t-50$

$\Rightarrow 10t=200$

$\therefore t=20\text{sec}$

Hence, the correct answer is option (d).