$A$ ball is projected horizontally with a speed $v$ from the top of an inclined plane which is inclined at an angle $45^{\circ}$ with the horizontal. How far from the point of projection will the ball strike the plane?

If the maximum range down an inclined plane is three times the maximum range up the same inclined plane,find the angle of inclination in degrees.

$A$ particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal (figure).
$(a)$ Find an expression for the range on the plane surface (distance on the plane from the point of projection at which the particle will hit the surface).
$(b)$ Find the time of flight.
$(c)$ Find the angle $\beta$ at which the range will be maximum.

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

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