A particle is project d from point A with velocity u at an angle `alpha` with horizontal. In its parabolic path, at point P the particle is moving at right angles to its initial direction of projection. Its velocity at P is:

Options

A. `u tan alpha`
B. `u cot alpha`
C. `u cos alpha`
D. `u cosec alpha`

Answer

b

Explanation

`vecu=u cos alpha hati+(u sin alpha)hatj`
`vecv=u cos alpha hati+(-v cos alpha)hatj`

`vecu.vecv=u^(2) cos^(2)alpha-u v sin alpha cos alpha`
`0=u^(2) cos^(2) alpha= u v sin alpha cos alpha`
`v=u cot alpha`