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

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A lawn sprinkler stands on level ground, $6.0\ \text{m}$ from a long, flat vertical wall that is tall enough for no water to pass over it. Its nozzle is at ground level. It throws water at $10\ \text{m/s}$ in every direction above the ground: towards every point of the compass, and at every angle of elevation from $0^\circ$ to $90^\circ$ , as suggested in the figure. Ignore air resistance and the size of the nozzle, and take $g = 10\ \text{m/s}^2$ .

Side view. A sprinkler at ground level throws water jets at 10 m/s at several angles of elevation, in curved paths. A long vertical wall stands 6.0 m from the sprinkler, and some of the jets end on the wall.

Let the water leave at speed $u$ . Show that a point at horizontal distance $r$ from the nozzle and height $y$ above the ground can be reached only if

\displaystyle y \le \frac{u^2}{2g} - \frac{g r^2}{2u^2}

What is the highest point on the wall that gets wet? At what angle of elevation is that water thrown, and how fast is it moving when it hits the wall?
The wet patch on the wall reaches down to the ground. Along what length of the foot of the wall is the wall wet?
Find the area of the wet patch on the wall.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the boundary of all the points the water can reach (the envelope of all the paths). Then cut it with the plane of the wall.

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

\displaystyle y = r\tan\theta - \frac{g r^2}{2u^2}\,(1 + \tan^2\theta)

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

\displaystyle \frac{g r^2}{2u^2}\tan^2\theta - r\tan\theta + \left(y + \frac{g r^2}{2u^2}\right) = 0

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

\displaystyle y_{\max} = \frac{(10\ \text{m/s})^2}{2 \times 10\ \text{m/s}^2} - \frac{10\ \text{m/s}^2 \times (6.0\ \text{m})^2}{2 \times (10\ \text{m/s})^2} = 5.0\ \text{m} - 1.8\ \text{m} = 3.2\ \text{m}

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

\displaystyle y(z) = \frac{u^2}{2g} - \frac{g(d^2 + z^2)}{2u^2} = 3.2\ \text{m} - \frac{z^2}{10 \times 2\ \text{m}}

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:

\displaystyle A = \tfrac23 \times 16.0\ \text{m} \times 3.2\ \text{m} = 34.1\ \text{m}^2

Answer: (b) $3.2\ \text{m}$ up, thrown at $59.0^\circ$ , arriving at $6.0\ \text{m/s}$ . (c) $16.0\ \text{m}$ . (d) $34.1\ \text{m}^2$ .

Check. The jet thrown at $59.0^\circ$ reaches the top of its path at a horizontal distance $\dfrac{u^2\sin 2\theta}{2g} = 4.41\ \text{m}$ , before the wall. So it is already falling when it touches the wall, and its speed there, $6.0\ \text{m/s}$ , is more than its horizontal part. A jet that just touches the boundary always meets it on the way down beyond its top, so this fits.

Watch out. Water that hits the wall stops there, so nothing reaches behind it. But each point on the wall is reached by the jet aimed at it, before that jet meets anything else, so the boundary on the wall is just the cut through the envelope.

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