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I hope this doesn't confuse you, but in one sense, yes, heavier bodies do fall faster than light ones, even in a vacuum. Previous answers are correct in pointing out that if you double the mass of the falling object, the attraction between it and the earth doubles, but since it is twice as massive its acceleration is unchanged. This, however, is true in the ...


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Although Pranav Hosangadi has explained, I will try to explain where you might be going wrong. I think this will be helpful for you. And I think it will not be waste of time in typing the answer for you. I was going to ask: if mass is an objects tendency to resist acceleration then why do two objects of different masses fall to the Earth at the same ...


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Mass is an object's tendency to resist acceleration. This applies when both masses you're testing are subjected to identical forces. From Newton's Law of Gravity, $$F = G \frac{M \cdot m}{r^2}$$ It is fairly obvious that the force the Earth exerts on a heavy body is more that what it exerts on a light body, so you can not compare the accelerations ...


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The velocity decreases at $(9.81 m/sec)/sec$ Divide that into the starting velocity to find the time it takes the velocity to get to zero.


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Assume air resistance is 0 and that the bullet goes completely vertically. I don't see any reason to model it as a quadratic. Just normal kinematic equations.


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I considered a similar question once. Consider a hollow (perfectly rigid) container of mass $M$ in a homogeneous gravitational field filled with $N$ identical particles of mass $m$ whirling around inside the container as an ideal gas in equilibrium (only exposed to the external gravitational field). Let the container be in contact with a weighing machine. ...


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This is only a partial answer but for gravitationally bound objects, i.e. the kinetic + potential energy < 0, you can simply use the equation you already have. If you start at a distance $y$ with velocity $u$ simply find the distance $y'$ where the velocity would be zero: $$ \frac{GM}{y_0^2} - \frac{GM}{y'^2} = \tfrac{1}{2}m u^2 $$ Then use this value ...


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For a $d=1\mu$m solid particle of density $1\text{g/cm}^3$, the very approximate terminal velocity according to the drag equation is (in Mathematica): << PhysicalConstants` r = 0.5 Micro Meter; \[Rho]s = 1 Gram/(Centi Meter)^3; m = 4/3 \[Pi] r^3 \[Rho]s; \[Rho] = 1.2 Kilo Gram/Meter^3; A = \[Pi] r^2; Cd = 0.47; g = AccelerationDueToGravity; ...


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What happens with very small spherical objects (d=1μm, e.g. a bacterium) in air? Do they fall? Yes they will fall. Acceleration of falling object depends on air resistance (And mass of an earth (at that case)). Falling object's acceleration doesn't depend on mass of bacteria (it depends on air resistance). Proof: According Newton's universal law of ...



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