The equations of motion of a projectile are given by $x = 36t \, m$ and $2y = 96t - 9.8t^2 \, m$. The angle of projection is

$A$ bullet is fired from a cannon with a velocity of $500 \, m/s$. If the angle of projection is $15^\circ$ and $g = 10 \, m/s^2$,then the horizontal range is:

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

$A$ ball $A$ is projected vertically upwards with a certain initial speed $u$. Another ball $B$ of the same mass is projected at an angle of $30^{\circ}$ with the vertical with the same initial speed $u$. At the highest point,the ratio of the potential energy of ball $A$ to that of ball $B$ will be: $(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2})$

The equations of motion of a projectile are given by $x = 36t$ metre and $2y = 96t - 9.8t^2$ metre. The angle of projection is:

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