A lawn sprinkler stands on level ground, 6.0 m from a long, flat vertical wall that is tall enough…

Step 1. The boundary (a). A jet thrown at angle $\theta$ follows

For a given point $(r, y)$ this is a quadratic in $\tan\theta$:

A real angle exists only if $r^2 \ge 4\,\dfrac{g r^2}{2u^2}\left(y + \dfrac{g r^2}{2u^2}\right)$, which gives $y \le \dfrac{u^2}{2g} - \dfrac{g r^2}{2u^2}$. On the boundary the two angles become one, with $\tan\theta = \dfrac{u^2}{g r}$.

Step 2. The highest wet point (b). On the wall, the lowest horizontal distance is $r = d = 6.0\ \text{m}$, straight in front of the sprinkler. There

This water is thrown at $\tan\theta = \dfrac{u^2}{gd} = 1.667$, so $\theta = 59.0^\circ$. By energy, its speed at the wall is $\sqrt{u^2 - 2g y} = \sqrt{100 - 64}\ \text{m/s} = 6.0\ \text{m/s}$.

Step 3. Along the foot (c). A point on the wall at a sideways distance $z$ from the foot of the perpendicular is at $r^2 = d^2 + z^2$. The boundary reaches the ground ($y = 0$) at $r = \dfrac{u^2}{g} = 10\ \text{m}$, the greatest range. So $z^2 = 10^2 - 6^2$, $z = 8.0\ \text{m}$ on each side, a length of $16.0\ \text{m}$ in all.

Step 4. The area (d). On the wall the wet patch lies under the curve

a parabola with its top at $3.2\ \text{m}$ and its base $16.0\ \text{m}$ long. The area under a parabolic arch is $\tfrac23 \times$ base $\times$ height: