# Tag Info

32

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 ...

12

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

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

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 ...

9

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 ...

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

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 ...

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 ...

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 ...

5

Firstly, that Wikipedia "Trajectory Calculation" page is pretty disappointing, it's not a very good fit to how smallarms ballistics is modelled and solved. A good book on the subject is Bryan Litz's recent Applied Ballistics for Long Range Shooting and a website with some first-rate on-line ballistics calculators as well as some good a very good writeups is ...

5

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.

5

Depends how much fuel you have. The space shuttle and other craft need such large heat shields because they use it to dump all their kinetic energy 'for free'. The shuttle only uses it's engines for takeoff and has only some small thrusters available in orbit. If you also had some form of propulsion as well as a space suit and could generate all the thrust ...

5

While writing out my progress on the problem, I managed to give myself the answer. So, I thought that I may as well share the solution as I have seen many people in my class get stuck here. If I have a kinetic energy equal to $K = (1/2)mv^2$ And I later have a velocity equal to half the original $v$ What happens to $K$? Shouldn't it be 1/4th the original? ...

4

In order to answer this question we must make some more assumptions. Given your scenario, one of the tragic school shootings which occurred in the States, we can assume the shooter will be within 50 feet. Since we are at a school, then let us assume typical materials found there such as desks and tables composed of laminate or other glued wooden materials ...

4

The drag on the projectile is determined by it's speed relative to the air around it. At high speeds air resistance goes as roughly the square of the air speed so it varies quite quickly with velocity. So if there is no wind the drag is proportional to the square of the projectile velocity. If there is a 5 m/s wind in the direction of travel the drag is ...

4

Consider the environment in which the propellent burns in a firearm. It is cramped space formerly packed tightly with stuff (the propellent, any necessary wadding and the bullet itself). There is damn little room for any atmosphere at all. Where---especially in a cartridge system---do you think the oxidizer (NB: not necessarily oxygen!) is coming from ...

4

UPDATE 2: If allowed to freely design the spacesuit the answer is yes.(see below) Initial answer before clarification The short answer is no. (SEE UPDATE) The longer answer is that there are a few key obstacles. One is that the person on the space station is in a relatively stable orbit around the planet. So while jumping from the space station would ...

4

Ok lets see. Differentiate the positions two times to arrive at the acceleration vector and see if it obeys Newtons Laws. $$\vec{r} = \begin{vmatrix} \left(v_0-\frac{C_DA\rho v_0^2}{2 m}t\right)\cos(\theta)t \\ \left(v_0-\frac{C_DA\rho v_0^2}{2 m}t\right)\sin(\theta)t-\frac{1}{2} g t^2+h \\ 0 \end{vmatrix}$$ $$\vec{a} = \ddot{\vec{r}} = ... 4 As ja72 points out, the formulas you have produced are, apart from a missing \frac{1}{2} factor, what you would get if drag was proportional to the square of the initial velocity, not the instantaneous one. For quadratic drag, you need to solve the following pair of equations:$$m \dot{v_x} = -k\sqrt{v_x^2 + v_y^2} v_x,m \dot{v_y} = -mg-k\sqrt{v_x^2 ...

4

If the ball has some spin, there's a differential drag force on either side of the ball, which is what causes "curveballs": the edge moving in the direction of the ball's travel "feels" the air moving past at a higher speed than the edge moving opposite the ball's direction of travel, so there's a greater aerodynamic drag on the fast side than the slow side. ...

4

The answer is the well-known Magnus effect which says that when a ball spins in a fluid it feels a force. A soccer ball's spin is most noticeable when it's spinning left-right (giving it the strong curve in either direction). This is the dominant kind of spin since soccer balls are kicked from the ground, making top-(or back-)spin difficult to achieve. This ...

3

Another one: A magician is locked in a wooden chest and is fired straight up from a very powerful cannon. He has a parachute to land safely, but it won't work if it's opened less than 25 meters from the ground. The magician tested the cannon (with a crash dummy, of course), and it can launch him up to 50 meters high. How much time will he have to escape ...

3

It's nothing to do with surface tension (art least for large objects). It's simply the force needed to accelerate the water out of the way to allow the object to sink. Imagine a bullet bouncing off another bullet, or metal armour. No problem accepting that, it's just Newton's laws and momentum. well water also has mass and needs a force to accelerate it in ...

3

If the initial velocity $v_0$ is allowed to be zero, here is one: You are standing next to a very deep hole, and wonder what is the depth $h$? You drop a stone into the hole, and hear a delayed sound (of the stone hitting the bottom) after time $T$ on your stopwatch. Given the gravitational constant $g$, given the speed of sound $c$, and by ignoring ...

3

Assuming you are ignoring air friction, then if $v_0$ is the muzzle speed of the bullet and $\theta$ is the angle from the horizontal, then the horizontal speed will be $v_h = v_0 cos(\theta)$ and the vertical speed will be $v_v = v_0 sin(\theta)$. So the bullet will reach a horizontal distance of $d = 2000 yards$ at a time $t = d/v_h$, so the time of flight ...

3

Marek asks for a "simpler conceptual argument"; there is one, using the basic observation that the distance travelled is linear in both the horizontal and vertical components of the initial velocity. It's not hard to fill in the rest, but I'm on a cross-country flight right now so it will need to wait a few hours. Edit Ok, details: The distance traveled ...

3

This types of problems are solved by observing projectile movements in $x$ and $y$ direction separately. In $x$ direction you have constant velocity movement $$v_x = v_{x0} = v_0 \cos(\theta), \; (1)$$ $$x = v_{x0} t +x_0 = v_0 \cos(\theta) \; t +x_0, \; (2)$$ and in $y$ direction you have constant acceleration movement with negative acceleration $-g$ ...

3

Air effects the projectile by the drag force, which is $$F_D = \frac{1}{2} \rho v^2 C_d A.$$ Here $\rho$ is density of air, $C_d$ is drag coefficient, $A$ is the cross-section area of the projectile and most importantly $v$ is relative velocity between air and projectile. So where does wind velocity $\vec{v}_a$ come into the equation? In a certain ...

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