If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

$A$ projectile is thrown with a velocity of $50 \, m/s$ at an angle of $53^o$ with the horizontal. Determine the instants at which the projectile is at the same 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 $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$

$A$ particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in a horizontal direction with a speed of $5 \; m/s$ from the same height. The correct statement is:

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

The maximum range of a gun on horizontal terrain is $16 \, km$. If $g = 10 \, m/s^2$,what must be the muzzle velocity of the shell in $m/s$?