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: In the following two problems we will look at shooting a basketball. Model the basketball as an elastic hollow sphere with radius 0.1 meters. Model the net and basket as shown below, dimensions marked. Neglect friction between the backboard and basketball, and assume all collisions are perfectly elastic.

<img_4294>


For this problem, you launch the basketball from the point that is 2 meters above the ground and 4 meters from the backboard as shown. You attempt to make a shot by hitting the basketball off the backboard as depicted above. What is the minimum initial speed required for the ball to make this shot?
Note: For this problem, you may assume that the size of the ball is negligible.

Solution:
For this part, we regard the basketball as a point mass. Now to account for the backboard bounce, we reflect the hoop over the backboard. The problem is now equivalent to making it into the hoop that is 4.15 meters away now. So now we need to find the smallest velocity to reach this point. One way to do it is to use the safety parabola, which makes it so minimal calculations are needed, so we present that method here. However, this problem is also readily solvable with only basic kinematics.
The point that the ball must hit (4.15 meters in front and 1 meter above) lies on the safety parabola when it is thrown at minimum speed. Since the focus of the safety parabola is the point of launch,



the directrix is

$$
\sqrt{1^{2}+4.15^{2}}=4.27 \mathrm{~m}
$$

away from the top of the hoop. The distance from the focus to the vertex of the parabola is then $(4.27+1) / 2=2.63 \mathrm{~m}$. So the minimum velocity to go this height is given by

$$
\frac{v^{2}}{2 g}=2.63 \mathrm{~m}
$$

and that gives $v=7.19 \mathrm{~m} / \mathrm{s}$. There is one last bit we need to check to complete this problem. The basketball's trajectory has to actually be possible (i.e. it doesn't go underneath the rim). It is clearly less optimal intuitively. If it passes through the underside, it means that its apex of the trajectory is very near its target.

Final answer: 7.19 m/s