Comments

Broofa, I agree with you. It really is a collision between the guy, the ball, and the Earth. (I did mention that in the last paragraph of my first comment but clearly should have included that in the original analysis!)Only for the horizontal component of the collision can we ignore Earth and gravity. In oversimplifying this into a one dimensional collision you rightly point out that we miss a major part of what is happening here. Including the Earth as part of the system we can still treat it as a collision using conservation of momentum, but as you suggest there is a greater change in momentum of the guy compared to the ball, which is compensated for by a change in momentum of the Earth (both linear and angular). Another more complicated way to look at the interaction would be by looking at each object individually and analyzing the external forces acting on these - which is what AreYouKidding is essentially suggesting. That is certainly a legitimate way to look at it. My problem there was with an important mistake he was making regarding the principle of conservation of momentum. Let's summarize! "In the two dimensional collision between the guy, the ball, and the Earth the total momentum of the system is conserved. The change in linear and angular momentum of "the guy" as a result of the collision is equal in magnitude to the change for the ball and Earth. The total change in momentum for the entire system is zero!" Thank you for your clarification/correction.

To AreYouKidding, Thank you for you comments, however you have made several errors in your critique that I have to point out. First of all you seem to be confusing the concept of kinetic energy with that of momentum. While it is true as you point out that kinetic energy is not conserved in inelastic collisons, momentum IS conserved. In the absence of external forces momentum is conserved in any and all collisions. Look that up in any introductory physics text book. Also I did not say that his rotation rate decreases as a result of his interaction with the ball, I said it increases. AFTER the collision, he is able to slow down his rotation once in the air specifically because conservation of angular momentum holds. It's very unclear to say that "compact objects require less momentum to rotate". What I think you mean is that compact objects tend to have less rotational intertia than extended ones making it easier to change their angular momentum with a given external torque. In the case of straightening out and slowing the spin rate the angular momentum of the guy is in fact constant (there us no external torque acting on him). Because he changes the distribution of mass he must slow down in order for momentum to be conserved. As far as energy being transformed from kinetic energy into elastic energy in the ball and then back to kinetic energy that is all true. It does not however contradict conservation of momentum in the collision. Now all that being said let's clarify an important point. Analyzing the system assuming no external forces is a simplification, and to be more accurate I should have included the Earth/ground as part of the system, because clearly the ground is exerting forces during the collison. If we acknowledge that, then any difference in changes in momenta of the guy and the ball would be accounted for by a change (basically impossible to detect) of the Earth itself.

titusrevised and rickyg001, The way it works is this. When a wave travels from one medium into another the wave speed does change. The wave speed is given by the equation v = freq. x wavelength. The change in speed is accomodated by a change in wavelength. However, the frequency always remains constant. Consider this example: the color of light depends on its frequency. If the frequency were to change in different media then the color would also change. However, if you look at an object under water its color is just the same as in the air. (Don't confuse this with dispersion of light which we see when light travels through a prism. In that case refraction separates white light into its individual frequencies.)

You guys may be right. It could well be a fake. It is interesting however that at least theoretically it should be possible. Of course the practical technical difficulties in construction might make it very difficult. Based on my own experience its technically quite difficult to launch several bottles simultaneously off the same pad. Anyway here's a quick and approximate calculation of the amount of thrust developed for a single half kg rocket which achieves a maximum altitude of 200 meters: Initial velocity after thrust: v = (2gy)^1/2 = 60 m/s Thrust force: F - mg = ma where a = delta v/delta t assuming a thrust time of 0.2 seconds we get F = 150N or so or about 35 pounds. With 25 rockets clearly that would be enough to lift a person. However whether that has ever really been done ...well maybe not.

Yikes! What a day. Yes your right about the conversion: 63 x 10 ^ 9/100 cm ^2 does equal 6.3 ^ 6. So ok yes if the cable is made of carbon nannotubes then it could be strong enough to withstand the impact and the weak link as you suggest could well be the lamp posts. As far as the centripetal force calculation it is correct. There is no special force called the centripetal force. The "centripetal" force is simply the vector sum of the forces acting radially on an object moving in a circle. So in this case the centripetal force is equal to the component of the air resistance force acting towards the center of his circular path minus the component of the gravity force acting away from the center of the circle. I took the easiest point of the motion to analyze which is at the bottom of the arc where the air friction component is straight up and the gravitational force acts straight down. If there is any horizontal component of force it is irrelevant to this calculation although it would add mean there is still a bit more force acting on Batman's arms. In the calculation therefore we are determining the force of air friction on Batman's arms which he in turn exerts back on the air.