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35

As other answers say, if someone just jumps off of the international space station(ISS), they would still be in orbit around the earth since the ISS is traveling at 17,000 miles per hour (at an altitude of 258 miles). Instead of just jumping, imagine the astronaut had a jet pack that could cancel that speed of 17,000 miles per hour in a very short time ...


19

This thread(physicsforums.com) contains a link to Shouryya Ray's poster, in which he presents his results. So the problem is to find the trajectory of a particle under influence of gravity and quadratic air resistance. The governing equations, as they appear on the poster: $$ \dot u(t) + \alpha u(t) \sqrt{u(t)^2+v(t)^2} = 0 \\ \dot v(t) + \alpha v(t) ...


17

Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes. Hence the speed and the kinetic energy of the falling body would both start to decrease after reaching the altitude where the terminal ...


16

It is indeed quite difficult to find information on why exactly this project has attracted so much attention. What I've pieced together from comments on various websites and some images (mainly this one) is that Shouryya Ray discovered the following constant of motion for projectile motion with quadratic drag: $$\frac{g^2}{2v_x^2} + \frac{\alpha ...


13

Let me first go through this without friction or air drag. You say $v_y$ along the $x$-axis and the train moves with $v_x$ along the $z$-axis. This is a little inconsistent. I will use the velocities, but not your description of the axes. So the train moves in the $x$-direction, the ball is thrown into the $y$-direction and it the $z$-direction is up-down. ...


13

In an ideal situation (no air resistance) there will be absolutely no difference in the place where the coin lands! Whether you toss the coin up from inside the train or while standing on the roof, the coin will land back in your hand (provided you've tossed it perfectly vertically). However, in practice, while standing on a fairly fast train's roof, ...


12

It depends on how you define the problem. Humans have re-entered the atmosphere from the International Space Station many times, by riding in either a Space Shuttle or a Soyuz capsule. Someone re-entering without a spacecraft of some sort would obviously have to wear some kind of pressure suit (as Felix Baumgartner did in his jump). How elaborate is the ...


11

45 degrees is, in fact, the angle for maximum range for a projectile with no air resistance. In the absence of air resistance, the only force acting is gravity, which causes a constant acceleration of g downwards. this determines the amount of time the particle spends in the air, via the formula for the position of a particle with constant acceleration: ...


11

The Red Bull Stratos project involving the 43-year-old Austrian man Felix Baumgartner is to break the sound barrier. Within the first 15,000 feet of his jump he was traveling well over the cruising speed of a commercial jetliner, reaching some 625 mph. The maximum velocity reached by Felix is about some 380 km/s. How did he do that? During a free-fall, ...


11

There's an interesting paper that discusses some of the physics/maths involved in the spiral path of a football. Here's Roberto Carlos' goal against France (discussed in comments to question). This is the way we interpret a famous goal by the Brazilian player Roberto Carlos against France in 1997. This free kick was shot from a distance of ...


10

The best intuition is a calculation but in this simple case, the calculation is really intuitive so you shouldn't turn off when you hear the word "calculation". The height reached by initial velocity $v$ is the height of the object after the initial velocity $v$ drops to $0$ (and then reverts the sign) because of the downward acceleration $g$. How much ...


10

If we are throwing two objects directly to the ground you are right. So from our kinematic equations: $$V_f = V_i + at$$ I would ask your teacher. What happens to the $V_f$ if $V_i=0$? Then Follow it up with what would $V_f$ be if $V_i$ was very large? The initial velocity DOES have an effect here. HOWEVER: Make sure that you are not misinterpreting ...


10

The answer: the ball appears to be deflected ~10 cm. The calculation: For simplicity, say we tee off at the north pole. The effects are a bit weaker at more typical locations, you multiply by sin(latitude) = 0.64 for a 40 degree (central california or washington DC) latitude. The Coriolis effect exists because the Earth rotates while the ball is in ...


10

tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ...


9

Yes your reasoning is correct,from the point of view of train the ball will travel in a tilted parabolic path as direction of apparent gravity will be different in the train and will not end up in your hand


9

At this stage, does the rocket still accelerate the craft? If by "velocity of the exhaust" we are talking about its velocity measured in the frame of the rocket, then Yes. Let $\mathbf u$ be the exhaust velocity as measured in the rocket frame, then in free space, the non-relativistic rocket equation is \begin{align} \frac{d\mathbf v}{dt} = ...


8

User Sahil Chadha has already answered the question, but here's the math and a pretty picture for anyone who is unconvinced that you're right. Since the train is accelerating, from the perspective of an observer on the train, the ball will experience a (fictitious) force in the direction opposite the train's travel having magnitude $ma$ where $m$ is the ...


8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...


8

It stays the same, because it's dependent on the square of velocity. You're probably wondering where the extra energy goes because potential energy is falling. The answer is: it is dissipated into heat because of friction from the air or air resistance.


7

At low velocities like this you can ignore special relativity and simply add the two velocities. This is really easy to see if you imagine yourself standing still and the Earth moving under you. Relative to you the gun should fire just like you were standing still. This is called an inertial frame of reference. You see the bullet leave at $400\: ...


7

The Wikipedia page on trebuchets links to a PDF paper which discusses exactly this question. It considers several models of varying complexity and finds a maximum range efficiency of 83% for a 100 pound counterweight, 1 pound projectile, a 5 foot long beam pivoted 1 foot from the point of attachment of the counterweight, and a 3.25 foot long sling. Here ...


7

My understanding is that at the speeds involved the shear forces inside the material are so tremendous that the target no longer behaves rigidly, but more like a liquid, which you can see in slow motion videos of bullets hitting metallic targets. Because the bullet is moving so fast, the "information" that the bullet has hit the target travels faster than ...


7

Well, this certainly is an evil trick to play on first year students! Escape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape "velocity" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from ...


7

If you solve for $t$ in Eq. (5.1), and plug that into equation (1.1), you'll see that the solution looks like $x_B \propto v_A^2 sin(\theta) cos(\theta)$. The function on the right is symmetric about $\pi/4$, thus, as long as $\theta$ doesn't equal $\pi/4$, there will be two solutions (symmetrically about $\pi/4$). Of course, in general, there could be ...


6

If there are no real solutions, that means that speed is too small to throw that rock so high! So the problem was never solved properly by the one who created it. IMHO, if you also have volume of the rock, you could calculate buoyancy and that would reduce gravitational acceleration and possibly rock could get so high.


6

The best way to prove something is wrong, is by performing a simple experiment, giving a counterexample. Take two identical objects (balls, pens, books). Throw one of the objects upwards and the other object downwards, so they have different initial velocities. The moment you let them go, they are in free fall. I am quite convinced the latter one will be one ...


6

Look thru the sniper scope carefully, and you may see not just a simple crosshair, but several horizontal lines intersecting the vertical line. Each of those is for a different distance. At short distances, the error is from the scope and rifle barrel being offset from each other. At long distances, the gravitaional drop of the bullet is also taken into ...


5

Bullets don't fly in straight lines. They fly in an arc, and often also veer sideways due to spin. Think how a pitcher throws a baseball. The fact that the scope is mounted above the barrel doesn't mean anything. Really, it could be mounted anywhere. Here is a good article on Wikipedia about it: Rifleman's Rule Picture says it all I think: What happens ...


5

I suspect your confusion is because you're holding two conflicting notions in your head, something like: "Motion is relative, so physics works exactly the same inside a moving train car as inside a stationary one." and: "Directions are absolute, so if I throw the coin straight up from the roof of the train, it goes the same way whether the train is ...


5

The time it takes is independent of the initial velocity if and only if the barrel is horizontal to the ground. If the barrel is horizontal, then the initial velocity will be solely in the $x$-direction and the only variable affecting the $y$-direction will be the acceleration due to gravity. However, if the barrel is at an angle, then the initial velocity ...



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