di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Mechanics

4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$
Context question:
(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$
Context answer:
\boxed{证明题}


Context question:
(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.

Plugging in this guess, what must $\lambda$ be?
Context answer:
\boxed{$\lambda=i \Omega$}


Question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.

This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.

Solution:
Using our answer to part (b) and $\alpha=A e^{i \phi}$, we have

$$
\eta(t)=A e^{i(\Omega t+\phi)}
\tag{25}
$$

Using the Euler identity, we have

$$
x(t)+i y(t)=A \cos (\Omega t+\phi)+i A \sin (\Omega t+\phi)
\tag{26}
$$

The real and imaginary parts become

$$
\begin{aligned}
& x(t)=A \cos (\Omega t+\phi) \\
& y(t)=A \sin (\Omega t+\phi)
\end{aligned}
\tag{27}
$$

Final answer: $x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$
Subfield: Electromagnetism


Question: (a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.

Solution:
In conductors, outer electron of each atom are free to move around the body. So if there is a non-zero electric field inside, this would move those free charges to the surface, at which point they cannot escape further, and they would accumulate there. This process would stop when the electric field created by the polarized charges exactly counter the external electric field inside. Consequently, any external electric field would cause a polarization of the body that results in a net zero field inside. Since $\nabla \phi=-\mathbf{E}=\mathbf{0}$, we have constant $\phi$ inside. From the continuity, it is also constant and equal to $\phi_{\text {inside }}$ throughout the surface.

Final answer: 开放性回答
Subfield: Electromagnetism

5. Polarization and Oscillation 

In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.

Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
Context question:
(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
Context answer:
开放性回答


Question: (b) Laplace's equation is a second order differential equation

$$
\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
\tag{8}
$$

Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.

Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.

Hint: Consider $\phi=\phi_{1}-\phi_{2}$.

Solution:
Suppose that we have two potentials satisfying the Laplace's equation and the boundary conditions. $\nabla^{2} \phi_{1}=0, \nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$. Define $\phi_{3}=\phi_{1}-\phi_{2}$. By the linearity of Laplace's equation, $\nabla^{2} \phi_{2}=0$ and $\phi_{3}=0$ on the surface. By the maximum principle, all extrema occurs at the surface. Thus, both the minimum and the maximum of $\phi_{3}$ is 0 . Therefore, $\phi_{3}=0$. This concludes the proof that $\phi_{1}=\phi_{2}$ identically.
Subfield: Electromagnetism


Question: (d) Argue from the uniqueness theorem that the electric field created by this image charge outside the sphere will be the same as the field created by the complicated polarization of the sphere.

Solution:
As argued in part (a), polarization of the sphere would create an equipotential surface. If we consider the region outside the sphere, the image charge configuration satisfies the necessary boundary conditions: The inner boundary (surface of the sphere) has equipotential with $\phi=0$, and the outer boundary (infinity) has zero potential. By the uniqueness theorem, potentials, and therefore electric fields, created by these different configurations are exactly the same throughout the volume in question.

Final answer: 开放性回答
Subfield: Electromagnetism

5. Polarization and Oscillation 

In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.

Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
Context question:
(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
Context answer:
开放性回答


Context question:
(b) Laplace's equation is a second order differential equation

$$
\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
\tag{8}
$$

Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.

Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.

Hint: Consider $\phi=\phi_{1}-\phi_{2}$.
Context answer:
\boxed{证明题}


Context question:
(c) The uniqueness theorem allows us to use "image" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?
Context answer:
\boxed{$x=\frac{R^{2}}{L}$ , $q_{0}=-q \frac{R}{L}$}


Context question:
(d) Argue from the uniqueness theorem that the electric field created by this image charge outside the sphere will be the same as the field created by the complicated polarization of the sphere.
Context answer:
开放性回答


Question: (e) Find the force of attraction between the charge and the sphere.

Solution:
From part (d) and (c), this is simply

$$
F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2} \frac{R}{L}}{\left(L-\frac{R^{2}}{L}\right)^{2}}
$$

Final answer: $F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2} \frac{R}{L}}{\left(L-\frac{R^{2}}{L}\right)^{2}}$
Subfield: Electromagnetism


Question: (f) Now suppose that we attach the point-like charge to a wall with a rod of length a. Any perturbation from the equilibrium will cause a perturbation of the polarization of the sphere. Prove that this equilibrium is stable and find the frequency of oscillation around it. The charge and rod have masses $m$ and $M$, respectively and assume that $L>R+a$.

![](https://cdn.mathpix.com/cropped/2023_12_21_78a48c63f6de68a072e3g-1.jpg?height=553&width=634&top_left_y=580&top_left_x=775)

Solution:
Force is attractive. Therefore, any deviation from the equilibrium $(\theta=0)$ induces an attractive torque towards $\theta=0$.

Force at the equilibrium would be the same as in part (e) with $L$ replaced by $L-a$. Since we are looking for small oscillations, we need to calculate the torque in the first order approximation of $\theta$. Denote $\beta$ as the angle between the line segment from the center of the sphere to the particle and from the particle to the wall. Then the equations of motion become

$$
\begin{aligned}
\tau=-F(\theta) a \sin (\beta) & \approx-F(0) a \frac{L}{L-a} \theta \\
& =I \frac{d^{2} \theta}{d t^{2}} \\
& =\left(m+\frac{M}{3}\right) a^{2} \frac{d^{2} \theta}{d t^{2}}
\end{aligned}
$$

Thus it follows that

$$
\omega=\frac{q}{(L-a)^{2}-R^{2}} \sqrt{\frac{R L}{4 \pi \epsilon_{0} a\left(m+\frac{M}{3}\right)}}
$$

In the limit $a<<L$, we obtain the formula

$$
\omega=\frac{q}{L^{2}-R^{2}} \sqrt{\frac{R L}{4 \pi \epsilon_{0} a\left(m+\frac{M}{3}\right)}}
$$

Final answer: 证明题
Subfield: Electromagnetism

<img_4389>

Figure 1: Cylindrical capacitor held above the fluid.

1. Dielectric Fluid 

Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor.

Question: (a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ?

Solution:
Gauss' law gives $E=\frac{\lambda}{2 \pi \epsilon_{0} r}$ inside the capacitor. The voltage is then determined by $V=-\int_{b}^{a} E d r$ which yields $V=\frac{\lambda}{2 \pi \epsilon_{0}} \log (b / a)$. The capacitance per length is determined by $\frac{\lambda}{V}=\frac{2 \pi \epsilon_{0}}{\log (b / a)}$, and the energy is found to be

$$
U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}
$$

Final answer: $U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}$
Subfield: Electromagnetism

<img_4389>

Figure 1: Cylindrical capacitor held above the fluid.

1. Dielectric Fluid 

Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor.
Context question:
(a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ?
Context answer:
\boxed{$U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}$}


Question: (b) How high $h$ does the dielectric fluid rise against the force of gravity given by acceleration $g$ ? (Note: it is very easy to get the correct answer while describing the problem incorrectly. Take care for full credit.)

Solution:
The key assumption is that $V$ is held constant, and therefore the capacitor system's energy balance is not isolated, and in general you should take into account the work done by the battery. The total energy is given by $U=U_{\text {bat }}+U_{\mathrm{EM}}+U_{\text {grav }} \cdot U_{\text {bat }}=-\int V d q=-V Q$ at constant $V, U_{\mathrm{EM}}=\frac{1}{2} Q V$ is the energy of the capacitor, and

$$
d U_{\text {grav }}=g y \cdot d m \Longrightarrow U_{\text {grav }}=\frac{\rho g y^{2} \pi\left(b^{2}-a^{2}\right)}{2}
$$



thus,

$$
U=-Q V+\frac{1}{2} Q V+\frac{\rho g y^{2} \pi\left(b^{2}-a^{2}\right)}{2}
$$

Equilibrium is found by minimizing this function, so

$$
0=-\frac{1}{2} \frac{d Q}{d y} V+\rho g y \pi\left(b^{2}-a^{2}\right)
$$

Now we must determine $\frac{d Q}{d y}$ as follows. From Gauss' law, $\int D \cdot d A=Q$. It's clear from the symmetry that the $E, D$ fields could only point radially, and that this automatically satisfies the boundary condition $D_{1}^{\perp}-D_{2}^{\perp}=0$ at the air-dielectric boundary. This symmetry allows us to break up the surface integral

$$
Q=\int_{\text {top part }} \epsilon_{0} E \cdot d A+\int_{\text {bottom part }} \kappa \epsilon_{0} E \cdot d A
$$

to conclude

$$
E=\frac{Q}{2 \pi \epsilon_{0} r[(\kappa-1) y+l]}
$$

which satisfies the remaining boundary condition. Then using $V=-\int E d r$,

$$
Q=\frac{2 \pi \epsilon_{0}[(\kappa-1) y+l]}{\log (b / a)} V
$$

and therefore

$$
\frac{d Q}{d y}=\frac{2 \pi \epsilon_{0}(\kappa-1)}{\log (b / a)} V
$$

and we conclude

$$
0=\frac{\pi \epsilon_{0}(1-\kappa)}{\log (b / a)} V^{2}+\rho g y \pi\left(b^{2}-a^{2}\right)
$$

which is solved to yield

$$
h=\frac{\epsilon_{0}(\kappa-1) V^{2}}{\log (b / a) \rho g\left(b^{2}-a^{2}\right)}
$$

Final answer: $h=\frac{\epsilon_{0}(\kappa-1) V^{2}}{\log (b / a) \rho g\left(b^{2}-a^{2}\right)}$
Subfield: Electromagnetism

<img_4389>

Figure 1: Cylindrical capacitor held above the fluid.

1. Dielectric Fluid 

Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor.
Context question:
(a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ?
Context answer:
\boxed{$U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}$}


Context question:
(b) How high $h$ does the dielectric fluid rise against the force of gravity given by acceleration $g$ ? (Note: it is very easy to get the correct answer while describing the problem incorrectly. Take care for full credit.)
Context answer:
\boxed{$h=\frac{\epsilon_{0}(\kappa-1) V^{2}}{\log (b / a) \rho g\left(b^{2}-a^{2}\right)}$}


Question: (c) Calculate the pressure difference $P$ above atmospheric pressure needed to suck the fluid to the top of the cylinder, assuming this is possible. Assume $l \gg h$ and neglect fluid dynamical and thermodynamic effects. You will find that $P$ consists of a term depending on $l$ and a term depending on $\kappa-1$. Provide a physical interpretation of these terms.

Solution:
Consider energy of the overall system. Define $A \equiv \frac{\pi \epsilon_{0} V^{2}}{\log (b / a)}$,

$$
\begin{aligned}
U_{i} & =\frac{\rho g h^{2} \pi\left(b^{2}-a^{2}\right)}{2}-A[(\kappa-1) h+l] \\
U_{f} & =\frac{\rho g l^{2} \pi\left(b^{2}-a^{2}\right)}{2}-A[(\kappa-1) l+l]
\end{aligned}
$$

so that

$$
W=\frac{\rho g\left(l^{2}-h^{2}\right) \pi\left(b^{2}-a^{2}\right)}{2}-A[(\kappa-1) l-(\kappa-1) h]
$$

This corresponds to work done by gas expanding on the liquid outside the capacitor with $W=\int P d V$ where $P$ is the gauge pressure. Then, assuming the pressure is constant with volume and neglecting terms of $O(h / l)$,

$$
P=\frac{\rho g l}{2}-\frac{\epsilon_{0} V^{2}(\kappa-1)}{\left(b^{2}-a^{2}\right) \log (b / a)}
$$

Final answer: $P=\frac{\rho g l}{2}-\frac{\epsilon_{0} V^{2}(\kappa-1)}{\left(b^{2}-a^{2}\right) \log (b / a)}$
Subfield: Mechanics

2. Trajectory of a point mass 

A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,

Question: (a) What value of $\theta$ maximizes the range?

Solution:
From the equation of motion $\frac{d \mathbf{v}}{d t}=\mathbf{g}$ we find that $\mathbf{v}=\mathbf{v}_{0}+\mathbf{g} t$. Then the position of the mass after time $t$ is given by $\mathbf{r}=\int_{0}^{t} \mathbf{v} d t=\int_{0}^{t}\left(\mathbf{v}_{0}+\mathbf{g} t\right) d t=\mathbf{v}_{0} t+\frac{1}{2} \mathbf{g} t^{2}$. Denoting the final time as $T$ and the position as $\mathbf{R}$, we find that

$$
\mathbf{R e}_{\mathbf{y}}=0=v_{0} \sin \theta T-\frac{1}{2} g T^{2}
\tag{1}
$$
$$
\mathbf{R e}_{\mathbf{x}}=L=v_{0} \cos \theta T
\tag{2}
$$

where $L$ is the range of the mass. From equations (1) and (2), we get

$$
L=\frac{2 v_{0}^{2}}{g} \sin \theta \cos \theta=\frac{v_{0}^{2}}{g} \sin (2 \theta)
$$

Maximum $L$ is achieved when $\sin (2 \theta)=1$. Therefore, $\theta=\frac{\pi}{4}$

Final answer: $\frac{\pi}{4}$
Subfield: Mechanics

2. Trajectory of a point mass 

A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
Context question:
(a) What value of $\theta$ maximizes the range?
Context answer:
\boxed{$\frac{\pi}{4}$}


Question: (b) What value of $\theta$ maximizes the surface area under the trajectory curve?

Solution:
The surface area covered by the trajectory can be found by

$$
\mathbf{A}=\frac{1}{2} \int \mathbf{r} \times \mathbf{d} \mathbf{r}=\frac{1}{2} \int(\mathbf{r} \times \mathbf{v}) d t
$$

Since the motion occurs in a plane, the surface area is always orthogonal to the plane. Therefore, the area covered by the trajectory can be found by calculating the magnitude of the vector $\mathbf{A}$. Now, inserting the equations for the position vector and the velocity as found in (1), we get

$$
\begin{aligned}
\mathbf{A}=\frac{1}{2} \int(\mathbf{r} \times \mathbf{v}) d t & =\frac{1}{4} \int\left(\mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) t^{2} d t \\
& =\frac{1}{4}\left(\mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) \int t^{2} d t \\
& \left.=\frac{1}{12} \mathbf{v}_{\mathbf{0}} \times \mathbf{g}\right) T^{3}
\end{aligned}
$$



As found in (1), $T$ is proportional to $\sin \theta$, and $\mathbf{v}_{\mathbf{0}} \times \mathbf{g}$ is proportional to $\cos \theta$. Therefore, the angle that maximizes the area maximizes the function

$$
f(\theta)=\cos \theta \sin ^{3} \theta
$$

Taking derivative of $f(\theta)$, we get the equation

$$
\frac{d f}{d \theta}=\sin ^{2} \theta\left(3 \cos ^{2} \theta-\sin ^{2} \theta\right)
$$

which gives zero at $\theta=0$ or $\theta=\arctan \sqrt{3}=\pi / 3$. Since $\theta=0$ corresponds to the minimal area (no trajectory, $\mathrm{A}=0$ ), the answer is $\theta=\pi / 3$

Final answer: $\pi / 3$ rads
Subfield: Mechanics

2. Trajectory of a point mass 

A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
Context question:
(a) What value of $\theta$ maximizes the range?
Context answer:
\boxed{$\frac{\pi}{4}$}


Context question:
(b) What value of $\theta$ maximizes the surface area under the trajectory curve?
Context answer:
\boxed{$\pi / 3$}


Question: (c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ?


Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows

$$
\frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v}
$$

Solution:
If the initial position of the mass is $\mathbf{r}_{\mathbf{0}}=h \mathbf{e}_{\mathbf{y}}$, equation (1) and (2) becomes

$$
\mathbf{R e}_{\mathbf{y}}=-h=v_{0} \sin \theta T-\frac{1}{2} g T^{2}
\tag{3}
$$
$$
\mathbf{R e}_{\mathbf{x}}=L=v_{0} \cos \theta T
\tag{4}
$$

Maximum $L$ occurs when $\frac{d L}{d \theta}=0$. Therefore, from (4), we obtain

$$
\frac{d L}{d \theta}=-v_{0} \sin \theta+v_{0} \cos \theta \frac{d T}{d \theta}=0
\tag{5}
$$

Taking derivative of both sides of (3) to find $\frac{d T}{d \theta}$ and inserting it into (5), we get

$$
\begin{aligned}
0 & =v_{0} \cos \theta T+\frac{v_{0} \sin \theta-g T}{\cos \theta} \sin \theta T \\
& \Longrightarrow v_{0}=g T \sin \theta
\end{aligned}
$$

Inserting this equation to (3), we obtain

$$
-h=\frac{v_{0}^{2}}{g}-\frac{1}{2} \frac{v_{0}^{2}}{g \sin \theta}
$$

Therefore, the answer is $\theta=\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$

Final answer: $\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$
Subfield: Mechanics

2. Trajectory of a point mass 

A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
Context question:
(a) What value of $\theta$ maximizes the range?
Context answer:
\boxed{$\frac{\pi}{4}$}


Context question:
(b) What value of $\theta$ maximizes the surface area under the trajectory curve?
Context answer:
\boxed{$\pi / 3$}


Context question:
(c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ?


Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows

$$
\frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v}
$$
Context answer:
\boxed{$\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$}


Question: (d) Supposing that $\beta<<\frac{g}{v_{0}}$, find the duration of the motion $T$.

Solution:
Under the presence of friction, equation of motion for $y$ component becomes

$$
\frac{d v_{y}}{d t}=-g-\beta v_{y}
\tag{6}
$$

Solving (6), we get

$$
v_{y}=v_{0 y}-\left(v_{0 y}+\frac{g}{\beta}\right)\left(1-e^{-\beta t}\right)
\tag{7}
$$



$y$ component of the final position vector can be found by integrating $v_{y}$,

$$
y=0=-\frac{g T}{\beta}+\frac{\left(v_{0 y}+\frac{g}{\beta}\right)}{\beta}\left(1-e^{-\beta T}\right)
\tag{8}
$$

Since $\beta$ is small compared to the dimensions of the system, we can use Taylor expansion for (8) to find an analytic equation for $T$.

$$
\begin{aligned}
& g T=\left(v_{0 y}+\frac{g}{\beta}\right)\left(1-e^{-\beta T}\right) \approx\left(v_{0 y}+\frac{g}{\beta}\right)\left(\beta T-\frac{1}{2} \beta^{2} T^{2}\right) \\
& \Longrightarrow T=\frac{2 v_{0 y}}{\beta v_{0 y}+g}
\end{aligned}
$$

Therefore, the answer is

$$
T=\frac{2 v_{0} \sin \theta}{g} \frac{1}{1+\frac{\beta v_{0} \sin \theta}{g}}
$$

which is less than the value found for frictionless trajectory, as expected.

Final answer: $T=\frac{2 v_{0} \sin \theta}{g} \frac{1}{1+\frac{\beta v_{0} \sin \theta}{g}}$