A freefalling body problem, only partial distance and time known

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.