If the kinetic energy of a second ball at its maximum height is $K$,what will be the kinetic energy of the first ball at its maximum height? (Assuming the first ball is thrown vertically upwards).

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 $m/s$?

The equation of a projectile is $y = \sqrt{3}x - \frac{x^2}{2}$. The velocity of projection is:

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:

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Two identical balls $A$ and $B$ thrown with the same velocity '$u$' at two different angles with the horizontal attain the same range $R$. If $A$ and $B$ reach maximum heights $h_{1}$ and $h_{2}$ respectively,then $R = 4 \sqrt{h_{1} h_{2}}$.
Reason $R$: The product of the said heights is $h_{1} h_{2} = \left(\frac{u^{2} \sin^{2} \theta}{2g}\right) \cdot \left(\frac{u^{2} \cos^{2} \theta}{2g}\right)$.
Choose the $CORRECT$ answer.

Two balls $A$ and $B$ are projected at an angle of $45^{\circ}$ and $60^{\circ}$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball $A$ to that for ball $B$ is $(\sin 30^{\circ} = \cos 60^{\circ} = 1/2, \sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}, \sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2)$.