A particle is thrown with a speed `u` at an angle `theta` with horizontal. After how much time the velocity of particle will be perpendicular to the initial motion of direction

Options

A. `(u)/(g cos theta)`
B. `(u)/(g sin theta)`
C. `(u sin theta)/(g)`
D. `(u cos theta)/(g)`

Answer

B

Explanation

Initial velocity `vec(u) - u cos theta hat(I) + u sin theta hat(j)`
After time `t`, velocity `vec(v) = u cos theta hat(i) + (u sin theta - g t) hat(j)`
`vec(u)` is perpendicular to `vec(v)`
`vec(u).vec(v) = u^(2) cos^(2) theta + u^(2) sin^(2) theta - g t u sin theta = 0`
`t = (u)/(g sin theta)`
OR

`x`: `v cos (90 - theta) = u cos theta` (i)
`y`: `v sin (90 - theta) = (u sin theta - g t)` (ii)
Dividing (i) by (ii), we get
`(sin theta)/(- cos theta) = (u cos theta)/(u sin theta - g t)`
`t = (u)/(g sin theta)`