A body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `(sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.

Explanation

At any height `y`, the speed of the projectile is
`v = sqrt(v_0^2 - 2 g y))`,
where `y = (y_(max))/(2)`
Sunstituting `y_(max) = (v_0^2 sin^2 theta_0)/(2 g)`, we have
`v = v_0 (sqrt(1 - (sin^2 theta_0)/(2)))`
Since `v = sqrt((2)/(3)) v_0` (given), we have `1 -(sin^2 theta_0)/(2) = (2)/(3)`
This given `sin theta_0 = sqrt((2)/(3))`. Hence, `theta_0 = sin^-1 (sqrt((2)/(3)))`.