# Derive ballistic equation using conservation of energy, stuck with (dx/dt)^2 [closed]

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 ===

• It looks like you have all of the physics down. This is just a math question – Aaron Stevens Oct 6 '19 at 15:35
• @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? – John Oct 6 '19 at 15:45
• 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. – Aaron Stevens Oct 6 '19 at 15:59