Q: Projectile Motion Tricky Problem 
A particle is projected from point A with a velocity u at an angle theta to the horizontal. At a certain Point B, it moves at right angle to its initial direction. what is the time taken from A to B? and what is the velocity at B?

Firstly I drew the diagram and found out the angle of the vector at B to the x axis, i.e. $90-\theta$.
So the horizontal components of velocity at $A$ and $B$ are $u\cos(\theta)$ and $v\cos(\theta)$. 
Since horizontal component is same we get $v=u\cot(\theta)$.
But how do we get the time taken from A to B?The answer in my book is $u/(g\sin(\theta))$.

A: I had encountered a similar question a few years back and the answer was like this-
$0=u-g\sin(\theta)t $
then-
$t=u/(g\sin(\theta))$
someone please explain what is meant by this.
