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When observing a falling particle, it's easy to derive the trajectory using the following reasoning: the particle is subject to one external force mg and by taking the double integral of the acceleration g we get

$y = -\frac{1}{2}gt^2 + v_0t + y_0t$

and with $v_0=0$ and $y_0=0$:

$y = -\frac{1}{2}gt^2$

I would like to get to the same result using the conservation of energy equation: potential energy U equals kinetic energy K

$K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{dy}{dt}\right)^2$

$U = mgh = -mgy$

and $K=U$

we can then write

$-2gy =\left(\frac{dy}{dt}\right)^2$ and then taking the square root, leading to

$\sqrt{-2gy} = \frac{dy}{dt}$ from there we rearrange to get

$\sqrt{-2g}dt = \frac{dy}{\sqrt{y}} = y^{-1/2}dy$

take the integral on both sides (for both t and y we start at 0)

$\int_0^t \sqrt{-2g}dt = \int_0^y y^{-1/2}dy$

$\sqrt{-2g}t = 2\sqrt{y}$

squaring on both sides

$-2gt^2 = 4y$

and hence

$y(t)= -\frac{1}{2}gt^2$

now, I really don't like this idea of taking the square root and then squaring the again and would prefer to work with $\left(\frac{dx}{dt}\right)^2$ but I don't know how to cope with the differentiator operator d, any ideas?

So basically at

$-2gy =\left(\frac{dy}{dt}\right)^2$ rewrite to get

$-2gdt^2 = \frac{dy^2}{y}$

=== this is where i am lost ===


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  • $\begingroup$ It looks like you have all of the physics down. This is just a math question $\endgroup$ – Aaron Stevens Oct 6 '19 at 15:35
  • $\begingroup$ @AaronStevens yes, I think the physics is correct but I would like help with that math step, should I post on the math stack exchange? $\endgroup$ – John Oct 6 '19 at 15:45
  • $\begingroup$ In my opinion, yes. But that's just my opinion. Do what you want. If the community here is fine with it then more power to you. $\endgroup$ – Aaron Stevens Oct 6 '19 at 15:59