# Projectile Motion Tricky Problem [closed]

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 $u/(g\sin(\theta))$.

## closed as off-topic by ACuriousMind♦, John Rennie, Kyle Kanos, jinawee, BMSOct 10 '14 at 19:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, John Rennie, Kyle Kanos, jinawee, BMS
If this question can be reworded to fit the rules in the help center, please edit the question.

• Its a good problem for first year physics. Here's a couple of things to ponder first. What if theta is 90? What if theta is zero? What if theta is 45? You should be able to logically answer these. Then test your solution against these conditions. – gogators Oct 10 '14 at 19:18

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.