Water Fountain: A Family of Trajectories

A fountain shoots streams of water at the same speed but different angles. Each stream is a perfect parabola — and together they form a stunning pattern bounded by an invisible curve called the safety parabola that no water can ever cross.

The Physics

Each stream is governed by the trajectory equation:

$y(x) = x\tan\theta - \dfrac{g\,x^{2}}{2 v_0^{2}\cos^{2}\theta}$

For a given $v_0$, varying $\theta$ from 0° to 90° gives an entire family of curves. Their outer envelope — the curve tangent to every member of the family — is itself a parabola called the safety parabola (or envelope of impact):

$y_{\text{safe}}(x) = \dfrac{v_0^{2}}{2g} - \dfrac{g\,x^{2}}{2 v_0^{2}}$

No droplet can ever reach above this curve. It's the protective dome of the fountain.

Step-by-Step Solution

Given:

• Jet velocity: $v_0 = 8\;\text{m/s}$
• Number of jets shown: 12
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The maximum range, the maximum height, and the equation of the safety dome.

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Step 1 — Find the maximum range across all angles.

The range as a function of angle is $R(\theta) = v_0^{2}\sin(2\theta)/g$, which is maximized when $\sin(2\theta) = 1$, i.e. at $\theta = 45°$:

$R_{\max} = \dfrac{v_0^{2}}{g} = \dfrac{(8)^{2}}{9.81} = \dfrac{64}{9.81} \approx 6.524\;\text{m}$

Step 2 — Find the maximum height across all angles.

Peak height $H(\theta) = v_0^{2}\sin^{2}\theta/(2g)$ is maximized when $\sin\theta = 1$, i.e. straight up at $\theta = 90°$:

$H_{\max} = \dfrac{v_0^{2}}{2g} = \dfrac{64}{19.62} \approx 3.262\;\text{m}$

(This is also the height of the safety dome at $x = 0$, directly above the fountain.)

Step 3 — Write the safety parabola explicitly.

Plug $v_0$ and $g$ into $y_{\text{safe}}(x) = v_0^{2}/(2g) - g\,x^{2}/(2 v_0^{2})$:

$y_{\text{safe}}(x) = \dfrac{64}{19.62} - \dfrac{9.81\,x^{2}}{128}$

$y_{\text{safe}}(x) \approx 3.262 - 0.0766\,x^{2}$

Step 4 — Find where the safety parabola hits the ground.

Set $y_{\text{safe}} = 0$ and solve for $x$:

$3.262 - 0.0766\,x^{2} = 0$

$x^{2} = \dfrac{3.262}{0.0766} \approx 42.59$

$x = \pm\sqrt{42.59} \approx \pm 6.524\;\text{m}$

That matches the maximum range — the safety parabola lands exactly where the 45° trajectory does. (And by symmetry, also $-6.524$ m on the left side.)

Step 5 — Compute one specific trajectory at $\theta = 45°$ to verify.

$v_{0x} = v_{0y} = 8\sin 45° \approx 5.657\;\text{m/s}$

$T = \dfrac{2 \times 5.657}{9.81} \approx 1.153\;\text{s}$

$H_{45} = \dfrac{(5.657)^{2}}{19.62} \approx 1.631\;\text{m}$

$R_{45} = (5.657)(1.153) \approx 6.523\;\text{m} \checkmark$

The 45° trajectory peaks at half the height of the safety dome and reaches the maximum range.

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Answer: The maximum range of any jet is $R_{\max} \approx 6.52\;\text{m}$ (achieved at $\theta = 45°$). The maximum height is $H_{\max} \approx 3.26\;\text{m}$ (achieved by the straight-up jet at $\theta = 90°$). The safety dome equation is:

$y_{\text{safe}}(x) \approx 3.262 - 0.0766\,x^{2}$

This parabola is tangent to every individual jet trajectory and meets the ground at $x = \pm 6.52\;\text{m}$.

Two Angles, One Range

Every range below the maximum is reached by two different angles — complementary angles that reflect about 45°. Throw at 30° and 60°, get the same range. The fountain visualization shows this beautifully when you have many jets.

Try It

• Increase the number of jets to see more trajectories fill in the pattern.
• Toggle the safety dome to see the envelope curve in green — every parabola is tangent to it but none cross.
• Bump up velocity and watch the whole pattern grow uniformly.
• Notice how the dome's height (above the fountain) is always half the maximum range. That's a quirky and beautiful geometric fact about parabolic trajectories.