$A$ 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?

$A$ 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.

$A$ man standing on a hill top projects a stone horizontally with speed $v_0$ as shown in the figure. Taking the coordinate system as given in the figure,the coordinates of the point where the stone will hit the hill surface are:

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.

$A$ small ball is projected up a smooth inclined plane with an initial speed of $10 \ m/s$ along a direction at $30^{\circ}$ to the bottom edge of the slope. It returns to the edge after $2 \ s$. The ball is in contact with the inclined plane throughout the process. Find the inclination angle $\theta$ of the plane. (in $^{\circ}$)

$A$ 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 ........ $\text{s}$.