A particle is projected from a point O with a velocity u in a direction making an angle α upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

The correct option is (u cosecα)/g.

To find the time of flight from the point of projection O to the point P where the particle's velocity is perpendicular to its initial velocity, we can analyze the motion using vector representation.

Let the horizontal direction be along the positive x-axis (represented by the unit vector $\hat{i}$) and the vertical direction be along the positive y-axis (represented by the unit vector $\hat{j}$).

The initial velocity vector $\overrightarrow{u}$ at point O, projected at an angle $\alpha$ to the horizontal, is given by:
$\overrightarrow{u}=u\cos \alpha \hat{i}+u\sin \alpha \hat{j}$

At any time t, the acceleration acting on the particle is due to gravity, which is directed vertically downwards: $\overrightarrow{a}=-g\hat{j}$.
Using the first equation of motion, $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$, the velocity vector $\overrightarrow{v}$ at point P after time t is:
$\overrightarrow{v}=\left(u\cos \alpha \right)\hat{i}+(u\sin \alpha -gt)\hat{j}$

Since the velocity vector at point P is perpendicular (at a right angle) to the initial velocity vector, their dot product must be equal to zero:
$\overrightarrow{u}·\overrightarrow{v}=0$

Substitute the expressions for $\overrightarrow{u}$ and $\overrightarrow{v}$ into the dot product equation:
$\left(u\cos \alpha \right)\left(u\cos \alpha \right)+\left(u\sin \alpha \right)(u\sin \alpha -gt)=0$

Expanding the terms, we get:
$u^2{\cos }^2\alpha +u^2{\sin }^2\alpha -ugt\sin \alpha =0$

Factoring out $u^2$ from the first two terms:
$u^2({\cos }^2\alpha +{\sin }^2\alpha )=ugt\sin \alpha$

Using the fundamental trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$:
$u^2=ugt\sin \alpha$

Dividing both sides by u (since $u\ne 0$):
$u=gt\sin \alpha$

Solving for the time t:
$t=\frac{u}{g\sin \alpha }$

Since the reciprocal of sine is cosecant ($\frac{1}{\sin \alpha }=\mathrm{cosec}\alpha$), we can write:
$t=\frac{u\mathrm{cosec}\alpha }{g}$

Thus, the time of flight from O to P is $\frac{u\mathrm{cosec}\alpha }{g}$.