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

The top of a smooth inclined plane of length $20 \sqrt{2} \,m$ is connected to the edge of a well of diameter $40 \,m$ making an angle $45^{\circ}$ with the vertical as shown in the figure. $A$ body is projected along the inclined plane with a velocity '$u$'. If the body crosses the well without falling into it, then the minimum value of '$u$' is $(g=10 \,ms^{-2})$.

$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is

If the range of a body projected with a velocity of $60 \,m \,s^{-1}$ is $180 \sqrt{3} \,m$, then the angle of projection of the body is (Acceleration due to gravity $= 10 \,m \,s^{-2}$)

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