Suppose a bullet were to be fired with a velocity $v$ from the surface at equator. Considering the spin of Earth will it land at the same spot?  I tried to solve this assuming constant acceleration due to gravity.  Since the horizontal velocity of the bullet remains constant at height $h$ the angular velocity will be in (angular velocity at ground is $\omega$ and radius of Earth is $R$) $$ \omega' =  \frac{\omega \times R}{(R+h)} $$ Now $\omega'$ can be written as $d\theta/dt$ and $h$ as  $v\times t-0.5\times g\times t^2$  so after rearranging we get $$ d\theta = \frac{\omega\times R}{(R+vt-0.5\times g\times t^2)}dt $$ Integrating this wrt to $t$ from $0$ to $2\times v/g$ gives me the angular displacement of the bullet. Angular displacement of ground is $2\times v\times \omega/g $and if we subtract the two and further multiply by $R$ we should get the distance traveled further by the ground.  However after calculating everything I'm getting very high distance. For example, a bullet fired at $1700m/s$ will land $2,340m$ away which seems absurdly wrong. Can anyone point out where my mistake is?
If I'm standing at the equator, jump, and land 1 second later, the  Earth does NOT move 1000mph (or .28 miles per second) relative to me,  since my velocity while jumping is also 1000mph.   However, the Earth is moving in a circle (albeit a very large one),  while I, while jumping, am moving in a straight line.   How much do I move relative to my starting point because of this? I  realize it will be a miniscule amount, and not noticeable in practise,  but I'd be interested in the theoretical answer.
MCMC in a frequentist setting I have been trying to get a sense of the different problems in frequentist settings where MCMC is used. I am familiar that MCMC (or Monte Carlo) is used in fitting GLMMs and in maybe Monte Carlo EM algorithms. Are there more frequentist problems where MCMC is used? Can a pixie fly higher than their altitude of one? In the pixie description on the compendium:     Altitude Limit: You fall at the end of your turn if you are using your racial fly speed and are more than 1 square above the ground (see the rules for flying and falling in the Rules Compendium).   So can a pixie choose to go two or more squares above the ground?   Here are the relevant rules from the compendium:  Fly speed:     A creature that has a fly speed can fly a number of squares up to that speed as a move action. If it is stunned or knocked prone while flying, it falls. See also “Flying” and "Hover".   Flying:     Flight      Some creatures have the innate ability to fly, whereas others gain the ability through powers, magic items, and the like. The rules for flight in the DUNGEONS &amp; DRAGONS game stress abstraction and simplicity over simulation. In real life, a flying creature’s ability to turn, the speed it must maintain to stay aloft, and other factors put a strict limit on flight. In the game, flying creatures face far fewer limitations.      FLIGHT      Flight follows the basic movement rules, with the following clarifications.         Fly Speed: To fly, a creature takes the walk, run, or charge action but uses its fly speed in place of its walking speed. A creature that has a fly speed can also shift and take other move actions, as appropriate, while flying.   Moving Up and Down: While flying, a creature can move straight up, straight down, or diagonally up or down. There is no additional cost for moving up or down.   Falling Prone: If a creature falls prone while it is flying, it falls. This means a flying creature falls when it becomes unconscious or suffers any other effect that knocks it prone. The creature isn’t actually prone until it lands and takes falling damage.   Remaining in the Air: A flying creature does not need to take any particular action to remain aloft; the creature is assumed to be flying as it fights, moves, and takes other actions. However, a flying creature falls the instant it is stunned, unless it can hover.   Landing: If a creature flies to a surface it can hold onto or rest on, the creature can land safely.   Terrain: Terrain on the ground does not affect a flying creature if the terrain isn’t tall enough to reach it. Because of this rule, flying creatures can easily bypass typical difficult terrain, such as a patch of ice on the ground. Aerial terrain can affect flying creatures.      Crashing        Falling while Flying: If a creature falls while it is flying, it descends the full distance of the fall but is likely to take less damage than a creature that can’t fly. Subtract the creature’s fly speed (in feet) from the distance of the fall, then figure out falling damage. If the difference is 0 or less, the creature lands without taking damage from the fall. For example, if a red dragon falls when it is 40 feet in the air, subtract its fly speed of 8 (8 squares = 40 feet) from its altitude. The difference is 0, so the dragon lands safely and is not prone.         If a creature is flying when it starts a high-altitude fall, it has one chance to halt the fall by making a DC 30 Athletics check as an immediate reaction, with a bonus to the check equal to the creature’s fly speed. On a success, the creature falls 100 feet and then stops falling. On a failure, the creature falls as normal.         High-Altitude Falls: Some encounters take place very high above the ground. In such an encounter, it is possible for a creature to spend more than one round falling to the ground. As a rule of thumb, a creature falls up to 500 feet during its first turn of falling. If it is still falling at the start of its turn, it can take actions on that turn as normal, then falls up to 500 feet at the end of the turn. If none of those actions expressly halts a fall, the creature falls up to 500 feet at the end of the turn. This sequence continues until the creature lands. Showing arrowheads in line end using QGIS? In Java, I'm dynamically creating a set of files and I'd like to change the file permissions on these files on a linux/unix file system.  I'd like to be able to execute the Java equivalent of chmod. Is that possible Java 5? If so, how?  I know in Java 6 the File object has setReadable()/setWritable() methods. I also know I could make a system call to do this, but I'd like to avoid that if possible. Placement of images in row using subfigure I want to put four figures in a single row. The code I am trying to use is taken from .  This puts three figures in a row. When I add a fourth one, I get three images in row and the last one appears in the second row. But I want all of them in one row. Here is my code.  \documentclass[12pt,twoside]{report}% Use this line for the print version of the thesis   \usepackage{graphicx} \usepackage{caption} \usepackage{subcaption} \begin{document}  \begin{figure}         \centering         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{gull}                 \caption{A gull}                 \label{fig:gull}         \end{subfigure}%           \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{gull}                 \caption{A gull2}                 \label{fig:gull2}         \end{subfigure}%          ~ %add desired spacing between images, e. g. ~, \quad, \qquad etc.           %(or a blank line to force the subfigure onto a new line)         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{tiger}                 \caption{A tiger}                 \label{fig:tiger}         \end{subfigure}         ~ %add desired spacing between images, e. g. ~, \quad, \qquad etc.           %(or a blank line to force the subfigure onto a new line)         \begin{subfigure}[b]{0.3\textwidth}                 \centering                 \includegraphics[width=\textwidth]{mouse}                 \caption{A mouse}                 \label{fig:mouse}         \end{subfigure}         \caption{Pictures of animals}\label{fig:animals} \end{figure} \end{document}   As I have mentioned above, it works for three figures.  But I can not use it for four figures. What should I change?   What does  width=\textwidth mean?   I changed    \begin{subfigure}[b]{0.3\textwidth}   to  \begin{subfigure}[b]{0.2\textwidth}   but it didn't help.  Then, I changed     \includegraphics[width=\textwidth]{mouse}   to  \includegraphics[width=0.5\textwidth]{mouse}   Still no solution. "He saw it stop and his son get out" I have been reading Men With Brooms for a few days. I read a sentence which did not make any sense to me, so I landed here to get some help on it. I have pasted the sentence from the novel. Please go through it and let me know your valuable feedback on it.  An excerpt from the novel (Men With Brooms: A Sweeping Epic, paperback 2002, a novelization by Diane Baker Mason):     Gordon was about to walk away from the Impala when he saw it stop and his son get out. So it was real the boy had come.   As per my opinion it should have been:     Gordon was about to walk away from the Impala when he saw it stopped and his son got out. So it was real the boy had come.   I have two questions here:   Why did the writer use stop and get instead of stopped and got? As she is telling us a story which happened in the past. Let's say the writer is telling us her mind's situation and she used the present tense for it but why did she use stop and get instead of stops and gets? Estimating partial sums $\sum_{n = 1}^m \frac{1}{\sqrt{n}}$ Apostol's Calculus, exercise number I 4.7 13.     Prove that if $n \geq 1$, then   $$ 2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1}) $$   and use this to prove that if $m \geq 2$, then   $$ 2 \sqrt{m} - 2 &lt; \sum_{n = 1}^m \frac{1}{\sqrt{n}} &lt; 2\sqrt{m} - 1 $$   In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.   After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:  assuming $2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:  $2(\sqrt{n+2} - \sqrt{n+1}) &lt; \frac{1}{\sqrt{n+1}} &lt; 2(\sqrt{n+1} - \sqrt{n})$  Any help is very appreciated, thanks "approach to address" or "approach to addressing"? Depending on our point of view, we offer different approaches to address(ing?) a same issue.   Should I use &mdash;ing in this context, and if yes, why? how to make clean without any modification? Answer appears automatically converted as a comment I posted an answer (pointing to another question already asked) but when I hit submit, I saw my answer appear as a comment to the question instead.   I tried it a couple of times and still the same result. The answer gets eaten up and is transformed into a comment.   Is this a bug? Bluetooth earphones work fine until sleep. After resuming from sleep however, they appear to connect for a brief moment before disconnecting. On blueman, the error given is Resource temporarily unavailable. This issue arose only after updating to 18.04 LTS.   Here's the terminal output for lsusb:  Bus 001 Device 002: ID 8087:8001 Intel Corp.  Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 003 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub Bus 002 Device 004: ID 1bcf:0002 Sunplus Innovation Technology Inc.  Bus 002 Device 003: ID 04f2:b477 Chicony Electronics Co., Ltd  Bus 002 Device 002: ID 0a5c:21f1 Broadcom Corp. HP Portable Bumble Bee Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub In previous Ubuntus (GNOME, not unity) I could add "--incognito" to the end of the executable file command of Chrome to make Incognito default.  I cannot find an executable file for Chrome in 12.04. How do I make Incognito default for Chrome in 12.04? SharePoint Migration 2007 to 2013 I'm doing some architectural cleanup that involves moving a bunch of classes into different projects and/or namespaces. Currently I'm moving the files by hand, building, and then manually adding using Foo statements as needed to resolve compilation errors. Anyone know of a smarter way of doing this? (We're a CodeRush and Refactor! shop, but I'd be interested to hear if Resharper has support for this) l, ls & la -- what are the differences, and are there more of these commands? Cross-referencing between different files How does one write on the outside of a circle?  I am looking for sample files/packages which imitate a magazine lay-out. Specifically a page with two columns having a circular picture between the two columns.    Example of the desired layout, created using Word: Review of a proof for De Moivre's theorem using mathematical induction The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$ regex to match word boundary beginning with special characters Folland, "Real Analysis", Chapter 5.3, Exercise 36.