$A$ 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 ....... $m$.

$A$ plane is inclined at an angle $\alpha = 30^o$ with respect to the horizontal. $A$ particle is projected with a speed $u = 2 \, m/s$ from the base of the plane,making an angle $\theta = 15^o$ with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to ........ $cm$ (Take $g = 10 \, m/s^2$)

$A$ 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:

$A$ 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 $g=10 \,m/s^2$]:

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 $\text{ m}$)

$A$ 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.