Step 1: Find the vertical height of the inclined plane.
The highest point of the inclined plane is at $8.75\,\text{m}$ from the ground.
The lowest edge of the roof is at $3.75\,\text{m}$ from the ground.
So the vertical height along the incline is:
\[
h = 8.75 - 3.75 = 5\,\text{m}
\]
Step 2: Find the speed at the lower edge using energy conservation.
For a sphere treated as a point mass (as per the official solution approach), all potential energy converts to kinetic energy:
\[
mgh = \frac{1}{2}mv^2 \implies v^2 = 2gh = 2(10)(5) = 100
\]
\[
v = 10\,\text{m/s}
\]
Step 3: Resolve the velocity at the launch point.
The sphere leaves the inclined plane at angle $30^\circ$ below the horizontal (since the incline makes $30^\circ$ with horizontal):
\[
u_x = v\cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\,\text{m/s} \quad (\text{horizontal})
\]
\[
u_y = v\sin 30^\circ = 10 \times \frac{1}{2} = 5\,\text{m/s} \quad (\text{downward})
\]
Step 4: Set up equations of projectile motion from the lower edge.
Taking downward as positive, the sphere must fall $3.75\,\text{m}$ vertically:
\[
s = u_y t + \frac{1}{2}g t^2
\]
\[
3.75 = 5t + \frac{1}{2}(10)t^2 = 5t + 5t^2
\]
Step 5: Solve for time.
\[
5t^2 + 5t - 3.75 = 0 \implies 4t^2 + 4t - 3 = 0
\]
\[
t = \frac{-4 + \sqrt{16 + 48}}{8} = \frac{-4 + 8}{8} = \frac{1}{2}\,\text{s}
\]
Step 6: Find the horizontal distance and state the answer.
\[
x = u_x \times t = 5\sqrt{3} \times \frac{1}{2} = \frac{5\sqrt{3}}{2}\,\text{m}
\]
\[
\boxed{\dfrac{5\sqrt{3}}{2}\,\text{m}}
\]