One second after the projection, a stone moves at an angle of ` 45^@` with the horizontal. Two seconds from the start,
it is travelling horizontally. Find the angle of projection with the horizontal. `(g=10 ms^(-2))`.

Options

A. `60^(@)`
B. `tan^(-1)(4)`
C. `tan^(-1)(3)`
D. `tan^(-1)(2)`

Explanation

(d) When stone travels horizontally then it must be at the maximum height.
So, `2=(u sin theta)/g`
`impliesu sin theta=20`……….(i)
Also, `tan 45^(@)=(u sin theta-g(1))/(u cos theta)`
`implies u cos theta=20-10`
`implies u cos theta =10`…………(ii)
Squaring of Eqs. (i) and (ii) and adding we get
`u=10sqrt(5)ms^(-1)` and `tan theta=2`
`implies theta=tan^(-1)(2)`