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

$A$ particle is projected horizontally from a tower with velocity $10\,m/s$. Taking $g=10\,m/s^2$,match the following two columns at time $t=1\,s$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

$A$ large number of bullets are fired in all directions with the same speed $u$. The maximum area on the ground on which the bullets will spread is

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.