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

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closed as off-topic by ACuriousMind, John Rennie, Kyle Kanos, jinawee, BMS Oct 10 '14 at 19:19

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  • $\begingroup$ 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. $\endgroup$ – gogators Oct 10 '14 at 19:18
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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.

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  • $\begingroup$ you are right.i think this is the way the time might have been calculated . someone please explain how this equation has come? $\endgroup$ – Jai Mahajan Oct 10 '14 at 17:25
  • $\begingroup$ how can we say that when the final velocity is zero it it the time taken to reach B? $\endgroup$ – Jai Mahajan Oct 10 '14 at 18:36

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