# Tag Info

## New answers tagged free-fall

0

You are not feeling the acceleration but you feel the forces acting on your body while accelerating . when your car is accelerating you are pressed against the back of the seat, when you are in a elevator which is moving up you feel the floor pushing against your feet harder. Now imagine you are sliding down a giant, steep slide in a park, you will ...

2

Although your question isn't posed very clearly I worked on the problem of non-spherical objects sliding/tumbling down a slope and don't mind sharing the following insight with you. Left, a cuboid and right, a sphere of comparable dimensions and mass. The inclination has been chosen deliberately high. Let's look at the forces and torques acting on both ...

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Drop a piece of paper and it glides sideways as well as flips. So aerodynamics (and hence the shape) affect the way things fall. Specifically aerodynamic forces have a center of pressure, which when ahead of the center of mass the body would rotate and flip, but if behind it will swing and stabilize at this orientation. This is the reason arrows, darts and ...

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You need to research the mechanics of drag and the Drag Co-efficient (start with this wiki page). A simple model (where drag is a Ram Pressure) holds the drag to be proportional to the square of the speed. This can be justified on simple momentum conservation grounds, in the case of pure ram pressure. The drag co-efficient is an empirically-found "fudge ...

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Your want to add a drag term to your force equation. It will be more complicated and involve the geometry of the objects in question. Note also your supposition will not always hold, an adult in a parachute will not fall faster than a child without a parachute. {disclaimer: Not tested empirically!!!!} You will have drag force term like: $$\vec{F}_{drag}\... 0 Yes it does. If you fix a (non inertial) frame of reference whose origin is on Earth's surface, at latitude \lambda, then a freely falling particle of mass m has the equation of motion,$$m\frac{d^2\vec r}{dt^2}=m\vec g_{ef}-2m\vec\omega\times\vec r,$$where$$\vec g_{ef}=\vec g-\vec\omega\times\left[\omega\times(\vec R+\vec r)\right]\approx \vec g-\vec\...

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Have a read through the answers to How can you accelerate without moving? and If $F=ma$, how do can we experience both gravity and a normal force even though we are not accelerating as these explain in some details exactly what is meant by acceleration in relativity. If you are falling freely then by definition you are weightless i.e. your proper ...

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