$A$ projectile is projected from the ground with an initial velocity $\vec{u} = u_0 \hat{i} + v_0 \hat{j}$. If the acceleration due to gravity $g$ is along the negative $y$-direction,find the maximum displacement in the $x$-direction (horizontal range).

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?

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.

$A$ particle is projected from a horizontal plane ($x-z$ plane) such that its velocity vector at time $t$ is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by:

$A$ bomb is dropped on an enemy post on the ground by an aeroplane flying horizontally with a velocity of $60 \ km/h$ at a height of $490 \ m$. At the time of dropping the bomb, the horizontal distance of the aeroplane from the enemy post so that the bomb hits the target is

Two projectiles are fired with the same initial speed from the same point on the ground at angles of $(45^{\circ}-\alpha)$ and $(45^{\circ}+\alpha)$,respectively,with the horizontal direction. The ratio of their maximum heights attained is: