Let us consider a ball is projected from $O$ at an angle of $45^{\circ}$ with horizontal and covering the path $O A B C$ (shown in figure).
(a) At point $B$, it will gain the same speed $u$ and after that speed increases and will be maximum just before reaching $C$. So greatest speed of ball is at $C$.
(b) During upward journey from $O$ to $A$ speed decreases and will be minimum at point $A$ where maximum height and $v_y=0$ has only horizontal component of speed $v_x$ $=\frac{u}{\sqrt{2}}$.
(c) As the force acting on ball is only due to gravitational force acts downward and constant so acceleration is always constant throughout the journey and is vertically downward equal to $g$.