Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A sprinkler fountain is in the shape of a semi-sphere that spews out water from all angles at a uniform speed $v$ such that without the presence of wind, the wetted region around the fountain forms a circle in the $X Y$ plane with the fountain centered on it.
Now suppose there is a constant wind blowing in a direction parallel to the ground such that the force acting on each water molecule is proportional to their weight. The wetted region forms the shape below where the fountain is placed at $(0,0)$. Determine the exit speed of water $v$ in meters per second. Round to two significant digits. All dimensions are in meters.

<img_4303>

Solution:
Method One: Let us look at only the major axis $x$. The wind provides a constant acceleration along $x$, which makes this problem equivalent to throwing a ball up and down a ramp tilted at an angle of $\beta=\tan \frac{a}{g}$. The furthest point along the $x$ axis represents the maximum distance you can throw an object up and down this ramp. This distance can be derived to be:

$$
d_{\mathrm{optimal}}=\frac{v^{2} / g}{1+\sin \beta}
\tag{6}
$$



where $\beta$ is negative if it's tilting downwards. We get the systems of equations:

$$

\frac{v^{2} / g_{\mathrm{eff}}}{1+\sin \beta}=0.25+0.25 \cdot \frac{13}{20}
\tag{7}
$$
$$
\frac{v^{2} / g_{\mathrm{eff}}}{1-\sin \beta}=1

\tag{8}
$$

where:

$$
g_{\mathrm{eff}}=g / \cos \beta
\tag{9}
$$

and solving gives $v=2.51 \mathrm{~m} \mathrm{~s}^{-1}$.

Method Two: Looking at the minor axis is much easier. The wind doesn't contribute in this direction, so we know the furthest point must have a vertical displacement of $d_{\max }=\frac{v^{2}}{g}$. We measure $d_{\max }=0.50+0.25 \cdot \frac{11}{20}$ and solving for $v$ gives $d_{\max }=2.50 \mathrm{~m}$.

Note that the answers aren't exactly the same due to measurement inaccuracies, so we asked for two significant digits. An earlier version of the question had the same diagram, but the labels were accidentally shifted. This did not affect the final answer.

Final answer: 2.5 m/s