Falling Body Equation: Derivation & Meaning

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The discussion centers on understanding the derivation of the equation of motion for a falling body under Earth's gravity. It begins with the expression md2s/dt2 = -mg, highlighting that the left side represents mass times acceleration, while the right side is the gravitational force. The cancellation of mass leads to the simplified equation d2s/dt2 = -g, indicating that the acceleration of the object is constant and equal to -g. The equation is a differential equation, requiring the determination of the function s(t) that satisfies it, resulting in the solution s(t) = -1/2 gt^2 + At + B, where A and B are arbitrary constants. Understanding this derivation provides insight into the physical implications of motion under gravity.
leehufford
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Hello,

I don't think I really understand the derivation of an equation of motion for a falling body under the force of Earth's gravity... I'm a little ashamed by this. I haven't been able to find someone asking quite the same question as I am about to ask.

I completely understand that the acceleration of an object is the second derivative of its position, and than an accelerated body is not in equilibrium, so ∑F ≠ 0.

I've been able to get by but it's time I truly understand what is going on. From the perspective of differential equations the derivation starts with an expression:

md2s/dt2 = -mg. Then masses cancel to get d2s/dt2 = -g.

What are we actually saying by equating two forces? I just see ma = -ma. But what does this statement physically mean?

I feel safe asking such a basic question on this forum and I trust that people will only give positive feedback in this situation. If I failed to find an online resource that explains this on the level I am seeking it would not be insulting to simply point me there. Thanks in advance,

Lee
 
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leehufford said:
From the perspective of differential equations the derivation starts with an expression:

md2s/dt2 = -mg. Then masses cancel to get d2s/dt2 = -g.

What are we actually saying by equating two forces?
Why do you think you are equating two forces? The only force is that of gravity, which equals -mg in your equation. The other side of the equation is mass times acceleration, from Newton's 2nd law.
 
leehufford said:
md2s/dt2 = -mg. Then masses cancel to get d2s/dt2 = -g.

What are we actually saying by equating two forces? I just see ma = -ma. But what does this statement physically mean?

In that equation, ##g## is a constant while ##s## is a function of ##t##, the time. The equation is telling you that the second derivative of ##s(t)## is equal to ##-g## and asking you to find the function ##s(t)## for which that is true.

That's what makes it a differential equation, it's an equation for an unknown function instead of just an unknown variable like you find in regular algebra.

The solution happens to be ##s(t)=\frac{-1}{2}gt^2+At+B## where ##A## and ##B## are arbitrary constants. It's a worthwhile exercise to plug this into the differential equation to verify that is in fact a solution... and then to figure out what the physical significance of the constants ##A## and ##B## is.
 
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