The horizontal range of a projectile projected at an angle of $45^{\circ}$ with the horizontal is $50 \ m$. The height of the projectile when its horizontal displacement is $20 \ m$ is (in $m$)

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.

$A$ body when projected at an angle $\theta$ with the horizontal reaches a maximum height $H$. The time of flight of the body will be ($g=$ acceleration due to gravity).

The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then