A particle I projected at an angle of elevation `alpha` after time `t` it makes an angle `beta`with horizontal. The velocity of projection is

Options

A. `(2g t cos beta)/(sin(alpha - beta))`
B. `(g t cos beta)/(sin(alpha - beta))`
C. `(g t )/(2 sin(alpha - beta))`
D. `(4g t )/(sin(alpha - beta))`

Answer

B

Explanation

`tan beta = (u sin alpha - g t)/(u cos theta)`
`(sin beta)/(cos beta) = (u sin alpha - g t)/(u cos alpha)`
`u cos alpha sin beta = u sin alpha cos beta beta - g t cos beta`
`g t cos beta = u sin (alpha - beta)`
`u = (g t cos beta)/(sin (alpha - beta))`