The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is

$A$ ball is projected in a manner such that its horizontal range is $n$ times the maximum height. Find the ratio of potential energy to kinetic energy at the maximum height.

$A$ stone is projected vertically upwards with speed '$v$'. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed '$v$'. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

Two projectiles,one fired from the surface of the Earth with velocity $10 \, m/s$ and the other fired from the surface of another planet with initial speed $5 \, m/s$,trace identical trajectories. The value of the acceleration due to gravity on the planet is ......... $m/s^2$.

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The angle with the horizontal at which the projectile was projected is:

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by: