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1

I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question. The air resistance can be written as : $$ R = \frac{1}{2}\,C_x\, \rho\, S\, v^2 $$ Note : The mass of the object is not in this equation. This is very important. Applying Newton's law to one of the object gives at any ...


0

A very crude estimate using the impact depth method: The density of the human body is almost the same as that of water, so you would expect that you'll lose most of your velocity after penetrating a depth equal to the width of your body (measured in the direction orthogonal to the contact area so if you go in head first, it will be the length of your body). ...


11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...


0

Counting just the deacceleration suffered, you can get an idea (the duration of the impact is related to the viscosity of the fluid). Using the speed you found, the impulse is: $$J=\Delta p=-(80\times31.3)=-2504\,N\cdot s$$ Estimating a duration for the impact with the water of for example $0.3\, s$, the average force suffered is: ...


19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...


44

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$ m g = k v^2 $$ And thus $$ v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...


2

Impact on water is a very complex topic. Your simple calculation just figures out the velocity of a free-falling body after a 50 m drop. That just tells you the initial relative velocity of body and water surface. It doesn't tell you much about the force at impact, or whether the person survived. There are two things that might kill on impact: high local ...


0

Skipping the (high-school level) math for a moment, and apply a layer of common sense gives us this: Objects attract each other, inertia depends on mass. Given the mass of the earth is large one can reasonably assume it to be stationary for human-scale purposes. If we are dropping a moon-sized object we may want to consider the earth's motion, but it won't ...


1

Newtonian mechanics should be accurate enough. There are general procedures that can help answer the question "is it safe to ignore this effect." See, for example, dmckee's comment. This answer is more of a question-specific approach. The force on the "falling" object by Earth is equal in magnitude to the force on the Earth by the falling object. This can ...


0

It would be safer to swing closer to the hill if I am understanding your question. I've never seen the movie you are talking about but, in general, a lesser height would provide a smaller time frame in which the jumpers would accelerate. A lower impact velocity would result in a higher chance of withstanding the crash.



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