\[\alpha\]
to the horizontal. The angle it makes with the horizontal will decrease until it becomes zero, then become negative. Ignoring air resistance, the motion of the projectile will be symmetric about the maximum height.
\[v_y =u sin \: \alpha -gt\]
.The horizontal component of velocity is
\[v_x = u cos \: \alpha\]
The angle the projectile makes with the horizontal is
\[tan \beta = \frac{u_y}{u_x} = \frac{u \: sin \: \alpha -gt}{u \: cos \: \alpha}\]
Rearranging this gives
\[t= \frac{u sin \: \alpha - u cos \: \alpha \: tan \beta}{g}\]