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According to Wikipedia, Galileo experimentally discovered that uniformly accelerated motion describes a parabolic trajectory using an inclined plane.

Today we are taught at school that a ball thrown up and forwards describes a parabola, and the proof is in the formula that describes the vertical component of distance, that is in the at^2+bt+c form, therefore the cartesian description of a parabola.

enter image description here

Does the 'locus' description of a parabola (points in space having the same distance to focus and directrix) have any physical meaning in this context?

Is there any meaningful physical quantity or vector that is equal when measured to focus and directrix while the ball is moving in the air?

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3 Answers 3

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The trajectory of a thrown object is secretly an ellipse, not a parabola, though the eccentricity of the "orbit" of a typical thrown object is so close to $e=1$ that it's hard to distinguish the two. At the other focus of the ellipse is Earth's center of mass; the focus of the parabola is the ellipse's empty focus.

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  • $\begingroup$ While it does not directly answer the questions, it is a good 'homework' answer. It corrects the assumptions in the OP and points to a direction of further study, i.e. Orbital mechanics $\endgroup$
    – simonpa71
    Sep 28, 2020 at 9:27
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The projectile parabola is:

$$y(x)=\tan(\varphi)\,x-\frac{1}{2}\frac{g}{v_0^2\,\cos^2(\varphi)}\,x^2$$

where $$ 0 \le x \le \frac{v_0^2\sin(2\,\varphi)}{g}$$

the focus of this parabola is :

$$p_f=\frac 12\,{\frac {{v_0}^{2} \left( \cos \left( \varphi \right) \right) ^{2 }}{g}}\tag 1 $$

thus the parabola $$y(x)\mapsto \tan(\varphi)\,x-\frac{1}{4\,p_f}\,x^2$$

you can see at this diagram the parabolas and the focus points for different angles $\varphi$ and a constant projectile velocity $v_0$

enter image description here

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I had this same question yesterday. The derivation is pretty hard to post, but... The directrix is the height that the projectile would reach if it was thrown straight up with the same initial velocity as in your diagram.

For the parabolic motion, the kinetic energy at the top of its flight is $\frac{m(v_{0 x})^2}{2}$. And it's height is $h_1 = \frac{v_{0y}^2}{2g}$ which comes from $mgh = \frac{mv_{0y}^2}{2}$.

If you could redirect the horizontal kinetic energy into vertical, gravitational potential energy it would add a new height of $h_2 = \frac{v_{0x}^2}{2g}$.

Add the two heights h1 + h2 and you get $\frac{v_{0x}^2 + v_{0y}^2}{2g}$ which is just $\frac{v_0^2}{2g}$. And $h=\frac{v_0^2}{2g}$ turns out to be the equation for the directrix if you do it out carefully.

The height of the focus of the parabola is the same as height when the projectile is traveling at 45° up or down. I guess I should have known this before now...the latus rectum of a parabola intersects the parabola where it has a slope of +/- 1 (or 45°).

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