For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be `sqrt(2)` times the maximum height reached by it. Show that the angle of projection is `tan^(-1)` (2) .

Explanation

Position vector of h (max point) form 0, is `(R)/(2) = (1)/(2) (u^(2) sin 2 theta)/(g)`
But ` R // 2 = sqrt(2) h _(max)` (given)
`:. (1)/(2) (u^(2) sin 2 theta)/(g) = (sqrt(2)u^(2) sin ^(2) theta)/(2g)`
`:. 2 sin theta cos theta = sqrt(2) sin theta * sin theta`
`rArr (2)/(sqrt(2)) = (sin theta)/(cos theta) rArr tan theta = sqrt(2)`
`:.` Angle of projection , `theta = tan^(-1) sqrt(2)`