$A$ projectile is fired with a velocity $v$ at a right angle to a slope which is inclined at an angle $\theta$ with the horizontal. The expression for the range $R$ along the incline is

If the maximum range of a projectile on horizontal ground is $6 \, km$,then the maximum range on an inclined plane with an angle of $30^o$ is ......... $km$.

The surface of a hill is inclined at an angle of $30^{\circ}$ to the horizontal. $A$ stone is thrown from the summit of the hill (point $A$) at an initial speed of $10 \text{ m/s}$ at an angle of $60^{\circ}$ to the vertical. If the stone strikes the hill at point $B$ as shown in the figure,the distance between $A$ and $B$ is (Take $g = 10 \text{ m/s}^2$) (in $\text{ m}$)

$A$ projectile is projected with speed $u$ at an angle of $60^o$ with the horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally,the range on the inclined plane will be:

$A$ particle falling vertically from a height hits a plane surface inclined to the horizontal at an angle $\theta$ with speed $v_0$ and rebounds elastically. Find the distance along the plane where it will hit the second time.

If the time taken by the projectile to reach point $P$ from point $Q$ is $T$,and the velocity at $P$ is perpendicular to the inclined plane,then $PQ$ =