A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

Answer

`(4)`

Explanation

Initial velocity
`u = u cos theta hat i+ u sin theta hat j`
Velocity after time `t` is
`v = u cos theta hat i+ (u sin theta - "gt") hat j`
Since `u and v` are perpendicular to each other, therefore,
`vec u. vec v = 0`
`(u cos theta hat i+ u sin theta hat j) xx [u cos theta hat i+(u sin theta - gt) hat j] = 0`
`u^2 cos^2 theta + u^2 sin^2 theta - u sin theta "gt" = 0`
`t = (u)/(g sin theta) = (20)/(10 xx (1)/(2)) = 4 s`.
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