A projectill is projected at an angle `(alpha gt 45^(@))` with an initial velocity u. The time t, at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is :-

Options

A. `t=u/g (cos alpha-sin alpha)`
B. `t=u/g(cos alpha+sin alpha)`
C. `t=u/g(sin alpha-cos alpha)`
D. `t=u/g (sin^(2) alpha-cos^(2) alpha)`

Answer

C

Explanation

`vec(v)=u cos alpha hat(i)+(u sin alpha-g t)hat(j) :' vec(v)=vec(v)_(x)=vec(v)_(y)`

`u cos alpha=u sin alpha-g trArr t=u/g (sin alpha-cos alpha)`