A ball of mass $160\, g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 \,m / s$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $\left(g=10\, m / s ^{2}\right)$ (in $kgm ^{2} / s$)

A body is projected at $t = 0$ with a velocity $10\,ms^{-1}$ at an angle of $60^o$ with the horizontal. The radius of curvature of its trajectory at $t=1\,s$ is $R.$ Neglecting air resistance and taking acceleration due to gravity $g = 10\,ms^{-2},$ the radius of $R$ is   ........ $m$

A particle is projected at angle $\theta$ with horizontal from ground. The slop $(m)$ of the trajectory of the particle varies with time $(t)$ as ...........

A projectile thrown with a speed $v$ at an angle $\theta $ has a range $R$ on the surface of earth. For same $v$ and $\theta $, its range on the surface of moon will be

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 same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$

Reason R: Product of said heights.

$h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)$

Choose the $CORRECT$ answer 

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