# A freefalling body problem, only partial distance and time known

Well, I've been trying to figure out a problem which I imposed on myself, so no literal values included. Unfortunately, my brain is not cooperating.

The problem states:

What is the height from which an object was dropped if it travelled the last $x$ units of distance in $t_x$ units of time?

Specific values and units are not important. It's just a thought problem, for the sake of it.

I was thinking along the lines:

The total height would be $h_0$ and total time would be $t_0$. The equation for the height can be obtained as an indefinite integration of the function $v(t)$ or through more common channels which do the same thing.

$h_0 = \frac{gt_0^2}{2}$ (1)

Right, that much is clear. This much is also true:

$h_0 = h + x$ (2)

$t_0 = t + t_x$ (3)

Which enables us to restate the equation (1) as:

$h+x= \frac{g}{2}(t+t_x)^2$

Now, we know the values of $x$ and $t_x$ and the value of gravitational acceleration, $g= ~9.80665$ $m/s^2$

All that remains is $h$ and $t$ and I just can't express it, everything I try to do doesn't give me an insight into their values. Is the system of equations under-constrained? I would really appreciate some insight, even if it is just to show the error of my ways.

I apologize for not consulting textbooks and other materials, I have the tendency to attempt to discover things on my own, if I can, of course.

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Hi Eulster, and welcome to Physics Stack Exchange! Good question :-) –  David Z Mar 11 '12 at 20:13

You need one more equation, preferably relating $h$ and $t$. Think about it. Which equation relates the two? Note that $h$ and $t$ are respectively height and time traveled in the time leading up to (not including) the time in which it travels the last $x$ units.

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The obvious one is $h = \frac{gt^2}{2}$, but that gives me nothing useful, unfortunately. –  Eulster Mar 11 '12 at 15:52
@Eulster That's exactly the one I was referring to.... Why not? Substitute the value for $h$ in those equations; solve the quadratic to get $t$. Now back-substitute to get $h$. Back substitute to get $h_0,t_0$. Done! –  Manishearth Mar 11 '12 at 15:55
Yes, there's the dreaded "solve the quadratic". I hate it when that pops up =P. –  Manishearth Mar 11 '12 at 15:55
Ahahah, I actually dismissed that one, I am uncertain why... Here it is, solved for $t$ and $h$: $t = \frac{2x}{g t_x} - t_x$ $h = \frac{g}{2}(\frac{2x}{g t_x} - t_x)^2$ –  Eulster Mar 11 '12 at 16:08
Happens to me as well ;p. Anyways, good luck with your forays into the exciting, unpredictable world of physics! :D –  Manishearth Mar 11 '12 at 16:19

I know it's been awhile and I don't know if OP cares, but here is my attempt to help out. We have $$h+x=\frac{g}{2}(t+t_x)^2$$ now inserting $$t=\pm \sqrt{\frac{2h}{g}}$$ for $t$ and squaring we get $$h+x=\frac{g}{2}\left( \frac{2h}{g}\pm 2t_x \sqrt{\frac{2h}{g}} +t_{x}^{2} \right)$$ Now $$x=\pm t_x \sqrt{2gh}+\frac{g t_{x}^{2}}{2}\implies \frac{1}{2t_{x}^{2}g}\left( x-\frac{gt_{x}^{2}}{2} \right)^2=h$$ Now $$h_0=h+x\implies h_0=\frac{x^2}{2t_{x}^{2}g}+\frac{x}{2}+\frac{t_{x}^{2}g}{8}$$ If I didn't trip over the algebra/basic arithmetic, I think this is what you want.

Hope this helps.

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Try not to give full solutions to homework questions, only hints. meta.physics.stackexchange.com/questions/714/… –  Manishearth May 11 '12 at 4:35
This is not a homework question. –  Ron Maimon May 11 '12 at 4:40
@RonMaimon: are you telling me that this is not typical of what freshman physics students typically get? –  Jerry Schirmer Aug 17 '12 at 4:33
@JerrySchirmer: Unfortunately, it's beyond the creativity level of textbook writers. It looks superficially like a homework question, but the constraints are unnatural, and are unlikely to have been stumbled in an elementary book. As typical a self-constructed problem, the solution isn't textbook clean. –  Ron Maimon Aug 17 '12 at 4:37

The reason you get confused is because you are thinking about this stuff all wrong, and you are making letter soup. These are the main problems for elementary students, and there are two simple philosophical shifts which will make these issues disappear.

First: in physics you are not looking for a lot of "equations" which relate the values in some made-up problem. This is what you do in physics classes, but it isn't the program of physics. In physics, you are looking for a complete picture of the motion! You want to know everything there is to know! In this case, you want to know where the particle is at all times. This is the first thing you find, and you write this down as

$$x(t) = {g t^2\over 2}$$

That's it for the physics, everything follows from this and mathematics, since this tells you the entire history of the fall.

You know two points on the trajectory:

$$h= {gt^2\over 2}$$ $$h+x = {g(t+t_x)^2\over 2}$$

You know g,x,t_x and you want to find h and t. In order to do this, you need to get rid of the stupid symbols as far as possible, by good choice of units (always, always do this, it shows you the idealized mathematical problem, and it is never taught in school, in fact, in school they tell you the opposite--- to keep the symbols around for dimensional consistency--- this is the opposite of teaching, it is teaching incompetence): In this case, set units of time and space so that g=2 and t_x=1, and you find

$$h = t^2$$ $$h+x = (t+1)^2$$

Now you want to solve for t and h. Multiply out the second equation, substituting the first relation

$$h+x = h + 2t + 1$$

and you find t:

$$t= {x-1\over 2}$$

and

$$h = {(x-1)^2\over 4}$$.

Next you make the dimensionally correct form, either by thinking about what your units must be, or else by redoing the problem with the silly extra letters around (now that you know what you are doing). The result is

$$h = {g\over 2} ( {x\over gt_x} - {t_x\over 2})^2$$

And because you solve the problem with constants set to 1 first, you can now see through the stupid symbols, the g and t_x, to the main relation. Always do this, and you will never be stumped by an elementary problem again.

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I would love to see you try and teach a class of freshmen this way. –  Jerry Schirmer Aug 17 '12 at 4:39
@JerrySchirmer: I only did a handful of lectures, never to undergrads, but I did regular problem sections on mechanics and electromagnetism years ago, and I always did things this way. It's usually reasonably effective to explain things correctly, although you have to prepare carefully because the notation will be different from class notes. I got pretty high marks from the students, at least in my memory, and one fellow (a very nice guy) said my section was most informative for his undergrad experience and said "thank you" (TA's don't get this a lot). Strangely enough, this was rewarding. –  Ron Maimon Aug 17 '12 at 4:53
... what it does do is turn your sympathies toward the students, away from the incompetent book and usually incompetent lecturing. So you become what the TA training course called a "union organizer" TA, where you are fighting on the side of your students against the evil and corrupt system. The training course warned in strong terms against this, but I never saw a real downside in this for the students, only to the TA. Your relationship with your educator colleagues does tend to suffer. So I didn't listen, and was always a conscious "union organizer". –  Ron Maimon Aug 17 '12 at 4:55