0
$\begingroup$

Is there an equation for the max height of a projectile, that includes initial height? All of the equations found thus far only have the initial velocity and angle. Below is the equation I am talking about:

$$h_{max}=\frac{v^{2}sin^{2}\theta}{2g}$$

Or, would I just add the initial height to the answer to the equation above? I was messing around with this Projectile Motion Simulator, and it does seem that adding the initial height to the equation above would yield the max height when such initial height is considered. Or at least with the set of cases I tested.

$\endgroup$
2
  • $\begingroup$ You are correct go with it! $\endgroup$
    – Qwerty
    Dec 15, 2023 at 16:31
  • $\begingroup$ Yep just add the initial height. $\endgroup$
    – whoisit
    Dec 15, 2023 at 16:36

2 Answers 2

0
$\begingroup$

Short answer:

For an approximation that assumes a vacuum and a flat earth, that equation is correct. You can just add the starting height to that maximum height to get the actual maximum.

Why:

The equations of motion that describe how an object moves in a vacuum and in a constant gravity field (flat earth) are linear differential equations. Because they are linear they obey something called "superposition*", which means that you can take two separate solutions and add them together.

In your case, one solution is "a rock sitting on a hill" and the other solution is "that same rock thrown at a certain velocity from altitude zero".

Disclaimer:

Note that, technically, this does not work on a spherical earth, or one with atmosphere, because both of those introduce nonlinearities. However, for low velocities you won't throw it far enough for the earth's curvature (or the changing gravity vector) to matter much at all, and for dense enough objects thrown slowly enough (i.e., rocks thrown by hand) neither will atmospheric drag.

Start lobbing feather dusters by hand and atmospheric drag will come to dominate. Lob artillery shells with modern guns, and I know that atmospheric drag will make a big difference, and I think the fact that the earth is round will start to affect your results significantly, too.


* Technically superposition holds if and only if the equations are linear, so the one implies the other.

$\endgroup$
2
  • $\begingroup$ So, what would be a cutoff for velocity or range that would still be respectably 'accurate' for adding the initial height? Most of the scenarios that I am calculating/simulating should be withing a kilometer in projectile range. $\endgroup$ Dec 16, 2023 at 4:24
  • $\begingroup$ There isn't a universal hard cutoff -- it depends on the problem at hand (range, radius of the body you're on, air density, the aerodynamics of your projectile) and the accuracy to which you want to make your prediction. Usually the cutoff for a particular problem is the point where the inaccuracy due to the math you're using becomes significant enough to seriously degrade the accuracy of the experiment/bombardment/whatever that you intend to undertake. It takes solving the nonlinear equations -- or at least finding upper limits on the errors -- to decide if you need to worry. $\endgroup$
    – TimWescott
    Dec 16, 2023 at 16:05
0
$\begingroup$

starting with

$$x=v\cos(\theta)\,t\quad, y=y_0+v\sin(\theta)-\frac 12 g\,t^2$$

from here you obtain

$$y-y_0={\frac {\sin \left( \theta \right) x}{\cos \left( \theta \right) }}-\frac 12\,{\frac {g{x}^{2}}{{v}^{2} \left( \cos \left( \theta \right) \right) ^{2}}}$$

thus $$ x_f=y-y_0=0=2\,{\frac {\sin \left( \theta \right) {v}^{2}\cos \left( \theta \right) }{g}} $$

and $~h_{\rm max}~$

$$y-y_0\bigg|_{x=x_f/2}=\frac 12 \frac{\sin^2(\theta)\,v^2}{g}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.