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 ......... $km$.

$A$ body is projected with an initial velocity of $10 \,m/s$ from the base of an inclined plane as shown in the figure. If the angle of projection with the horizontal is $60^{\circ}$ and the inclination of the plane is $30^{\circ}$, then the range $R$ along the inclined plane is [take $g=10 \,m/s^2$]:

$A$ particle is projected from a point $A$ with velocity $u\sqrt{2}$ at an angle of $45^{\circ}$ with the horizontal as shown in the figure. It strikes the plane $BC$ at right angles. The velocity of the particle at the time of collision is

$A$ particle is projected at an angle of $60^{\circ}$ with a speed of $10\sqrt{3} \text{ m/s}$ from point $A$ as shown in the figure. At the same time,the wedge is made to move with a speed of $10\sqrt{3} \text{ m/s}$ towards the right as shown in the figure. The time after which the particle will strike the wedge is ........ $\text{s}$.

$A$ projectile is fired with a velocity of $21 \, m/s$ at an angle of $\beta = 60^o$ with the horizontal,up an inclined plane making an angle of $\alpha = 30^o$ with the horizontal. The range $R$ on the incline is ....... $m$.

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 $PQ$ =