# Tag Info

## Hot answers tagged projectile

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 ...

25

Olin Lanthrop suggested a plausible approach but there was a lot of (inaccurate) guessing in his answer. I was going to write this as a comment to his answer but it got too long. Note - in the below I round to no more than 2 significant figures - the nature of the problem doesn't support more. Let's take the famous Sherman tank as our example. A brief ...

22

It will travel along a parabola (ignoring drag from the air here), initially with upward velocity, as you describe in your first scenario. You're correct that the only force acting on the box is its weight, but this means it will have downward acceleration immediately, not necessarily downward velocity. Eventually the downward acceleration will lead to ...

22

So let's start with your last question, informally, the radius of curvature is a measure of how much a certain curve is pointy and has sharp corners. Given a curve $y$, you can calculate its radius of curvature using this formula: $$\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^\dfrac{3}{2}}{\left|\dfrac{d^2y}{dx^2}\right|}$$ You might ask what ...

20

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: $$\dot u(t) + \alpha u(t) \sqrt{u(t)^2+v(t)^2} = 0 \\ \dot v(t) + \alpha v(t) ... 17 Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes. Hence the speed and the kinetic energy of the falling body would both start to decrease after reaching the altitude where the terminal ... 16 I took a look at the clip and my take on it that the object of the exercise was not to slow down the tank, but to move it sideways and land in the lake a half a mile away. Hannibal says, "rotate the main gun to 82º" which I take to be sticking out sideways. Background Information The facts I could find were: An M1A2 Abrams weighs 69.54 tons or 63085.63 ... 16 In an ideal situation (no air resistance) there will be absolutely no difference in the place where the coin lands! Whether you toss the coin up from inside the train or while standing on the roof, the coin will land back in your hand (provided you've tossed it perfectly vertically). However, in practice, while standing on a fairly fast train's roof, ... 16 It is indeed quite difficult to find information on why exactly this project has attracted so much attention. What I've pieced together from comments on various websites and some images (mainly this one) is that Shouryya Ray discovered the following constant of motion for projectile motion with quadratic drag:$$\frac{g^2}{2v_x^2} + \frac{\alpha ...

14

I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a ...

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. ...

12

Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground. Just consider the vertical force caused by the air friction: $F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$ Where $\theta$ is the angle above the horizon for the bullet's velocity, ...

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 ...

11

Do the math. I'm no military expert, so I'll guess at some parameters, but I think it will show the effect is so small that it doesn't matter even if the numbers were considerably more favorable. Let's say the tank weighs 50 tons, which I think is rather light for a tank. That's 100,000 pounds, which puts its mass at 45,000 kg. I don't know what the mass ...

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 ...

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

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 ...

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} = ...

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

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.

8

I would consider that since acceleration is a constant vector pointing downward, that the time the projectiles downward component takes to accelerate from V(initial) to 0 would be the same as the time it takes to accelerate the object from 0 to V(final)

7

Correct me if I'm wrong, I believe that from a paintball gun when shooting there is no smoke, vapour or other visible exhaust and not even recoil. No visible signal. The dodger must rely on sound. Sound reaching the ears. The speed of sound in air is around $v=330\:\mathrm{m/s}$. Sound will propagate the $x=8\:\mathrm{m}$ and reach the dodger in ...

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 ...

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