When two balls of different masses are dropped from equal height, they reach the ground at the same time. Can anyone explain this in terms of physics?
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Too localized. Now, it turns out to be a possible duplicate – Waffle's Crazy Peanut Jun 11 '13 at 18:04
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Also of http://physics.stackexchange.com/questions/5973 (I thought it would show the second link when I closed it, but apparently not) – David Z Jun 11 '13 at 20:21
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Newton's law says that the force $\vec F$ exercing on an object produces an acceleration $\vec a$ such as :
$$\vec F = m_I \vec a$$ where $m_i$ is the inertial mass of the object.
On the other side, in your experience, the force is the gravitationnal force (the weight) $\vec P$ which is
$\vec P = m_G \vec g$, where $m_G$ is the gravitational mass, and $\vec g$ is the gravity acceleration.
The equivalence principle says that the inertial mass and the gravitational mass are equal, so $m_G = m_I$.
You have $\vec F =\vec P$, that is $m_G \vec g = m_I \vec a$
But $m_G = m_I$, so the acceleration is $\vec a = \vec g$, and this does not depends on the mass.
Trimok
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This is because both experience the same acceleration, g(9.81 m/s2). The acceleration is independent of mass. Thus when they are dropped, they will always maintain the same velocity and travel the same distance (neglecting air resistance of course!) in the same time.
yankeefan11
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"neglecting air resistance of course!" and Archimedes' principle too ... some balls tend to float upward. – babou Jun 11 '13 at 23:21