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49

The same reason objects which are heavier on one side tend to fall with the heavy side down: the tip of the arrow is denser than the rest of the arrow. The center of gravity is offset from its geometrical center, so the air drag, which is based on the object's geometry, causes a torque together with gravity as seen in this very professional picture of a body ...


46

Air. Conservation of angular momentum does infact dictate that whatever rotation it starts with it should end with, provided nothing else acts on it. Air allows its forward momentum to act on it. Consider a weather vane, a wind sock, or a flag. They rotate when not facing into the wind because one side presents more wind resistance than the other. Once ...


12

In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum? An arrow shot on the Moon would not do that. Air and the geometry of the arrow are key. An arrow flying through the air is subject to two forces, gravity and aerodynamic drag. Gravitation will not make an ...


9

The zig-zag strategy seems trivially obvious - but it might not be the better strategy in a particular situation. I suggest rather than asking under what conditions this strategy is preferable, you ask under what conditions the counter-intuitive straight-line strategy is preferable. The advantage of zig-zagging is that it presents a smaller "collision ...


8

As pointed out by dmckee in his comment, anyone (including myself) that has practised bow and arrow knows the arrow by weight and fletch inspection. As my English corrector points out, fletch doesn't seem to be a very commom word: it means that feather at the end of the dart/arrow. Everything resumes to how good is the fletching. When shot the arrow wants, ...


2

If we are at the equator on the surface of an atmosphere-free, non-rotating, perfectly spherical planet (no mountains or trees, etc) then the optimal angle should be zero (launch towards the horizon, tangent to the planet's surface). This assumes the initial velocity is the orbital speed at the surface of the planet, which is approximately the escape ...


2

I'll have an attempt at answering this. Typically Balloons are made of latex, which has a Young's Modulus of $$ 0.1 \times 10^{9} GPa $$ Bullet weights 0.008kg (8g) Radius of bullet is 9mm. Speed of 759.968 m/s. It takes a deformation of 2.5cm to break a latex balloon. Using the equation for youngs modulus and calculating the work needed to pierce both ...


2

That depends on the frame of reference. With respect to the man who throws the object, it will travel at a speed equal to the speed at which it was thrown. With respect to the ground, it'll fall straight down.


1

Yes indeed. I can't resist answering with This clip from Mythbusters which demonstrates exactly what you're asking about.



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