What is the angle between the velocity and acceleration of a projectile at its maximum height (in $°$)?

The time of flight of an object projected with speed $20 \, m/s$ at an angle $30^{\circ}$ with the horizontal is .... $s$.

$A$ body projected at a certain angle $(\neq 90^{\circ})$ from the ground crosses a point in its path at a time $t_1 = 2.3 \ s$ and from there it reaches the ground after a further time $t_2 = 5.7 \ s$. The maximum height reached by the body is (Take $g = 10 \ ms^{-2}$) (in $m$)

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

$A$ projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$ with velocity $v_1$. At the same time,another projectile $B$ is thrown with velocity $v_2$ upwards from point $Q$,which is vertically below the highest point that $A$ would reach. For $B$ to collide with $A$,the ratio $\frac{v_2}{v_1}$ should be:

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)