66Let the angle be \alpha. If you choose a coordinate system whose x axis runs along the incline and the y axis perpendicular to the incline then in this coordinate system the acceleration due to gravity will have a component -g\cos\alpha along the y axis and -g\sin\alpha along the x axis. Accordingly the velocity components v_x and v_y could be obtained as:
v_x = u\cos(\theta -\alpha) -g t\sin\alpha and
v_y = u\sin(\theta -\alpha) -g t\cos\alpha
Here I have taken u = 10 m/s and \theta = 60^\circ.
The y coordinate :
y = ut\sin(\theta -\alpha) -\frac{1}{2}g t^2\cos\alpha
When it returns back to the incline y=0. Accordingly we get the time of flight above the incline as
T = \frac{2u\sin(\theta -\alpha)}{g\cos\alpha}
Going back to the normal xy system (x axis horizontal y axis vertical), the angle \beta which the velocity vector makes with the horizontal is given by
\tan \beta = \dfrac{u\sin \theta -gt}{u\cos \theta} = \tan\theta - \frac{gt}{u\cos\theta}
This is decreasing function of t. So we require that the least value of β (attained for t= T) should not be less than 30°. That gives
\tan\theta - \frac{gT}{u\cos\theta} \ge \frac{1}{\sqrt{3}}
Solving this inequality gives
\boxed{ \alpha \ge 30^\circ}