A particle is thrown with speed `u` at an angle `alpha` with horizontal from the ground. After how much time, the velocity of particle will make an angle `beta` with horizontal.

Explanation

The particle velocity will be directed at angle `beta` with horizontal after time `t_(1)` and `t_(2)` while ascending and descending respectively.
`O` to `A` :
`v_(x) = u_(x) = u cos alpha`
`v_(y) = u sin alpha - g t`
`tab beta = (v_(y))/(v_(x)) = (u sin alpha - g t_(1))/(u cos alpha)`
`(sin beta)/(cos beta) = (u sin alpha - g t_(1))/(u cos alpha)`
`u cos alpha sin beta = u sin alpha cos beta - g t_(1) cos beta`
`g t_(1) cos beta = u sin(alpha - beta)`
`t_(1) = (u sin (alpha - beta))/(g cos beta)`
`O` to `B` :
`tan beta = (v_(y))/(v_(x)) = -(u sin alpha - g t_(2))/(u cos alpha)` [`v_(y)` is along the negative y-axis]
`(sin beta)/(cos beta) = (- u sin alpha + g t_(2))/(u cos alpha)`
`t_(2) = (u sin (alpha - beta))/(g cos beta)`
`[[sin(alpha+beta)=sin alpha cos beta+cos alpha sin beta,],[sin(alpha-beta)=sin alpha cos beta-cos alphasin beta,]]`