A particle is projected from a point P with a velocity v at an angle `theta` with horizontal. At a certain point Q it moves at right angles to its initial direction. Then

Options

A. Velocity of particle at Q is `v sin theta`
B. Velocity of particle at Q is `v cot theta`
C. Time of flight from P to Q is `(v//g) cosec theta`
D. Time of flight from P to Q is `(v//g) sec theta`

Answer

B, C

Explanation

`:'` Horizontal component of velocity remains constant
`:. v sin theta=v cos theta` (from figure) `:. v'=v cot theta`

So from `v_(y)=u_(y)+a_(y)trarr -v' cos theta`
`=sintheta-g t-v(cos^(2)theta)/(sin theta)=v sintheta-g t :. t=v/g cosec theta`