The general notion might be from physics, but I am very much interested in the math core of the problem. I've been trying to figure it out for a while, with little success.

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. So, we don't have any literals, but what we do have is:

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)

$h = \frac{gt^2}{2}$

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.

So, the point of all this is, how can I obtain the original height function just from knowing two facts, the length traveled and the time it took to do it.

Can a function be reconstructed? And how would one go about doing it? Without succumbing to averaging the end velocity (stating $x/t_x$) and then figuring out how it "decayed" backwards in time, therefore reaching an approximation of the height.


1 Answer 1





$$h+x=\frac g2(t+t_x)^2$$


$$x=gtt_x+\frac g2t_x^2\;.$$

You can solve this for $t$ and then determine all the other variables from that.

  • $\begingroup$ Actually, that's a brilliant approach. Both sides are downsized by an equal value (as inferred by = ) and solved for a target variable. Thank you. $\endgroup$ Mar 11, 2012 at 18:42
  • $\begingroup$ But, if I may ask, why does this work and substitution doesn't? I tend to zero out the targeted variables when I use $h = gt^2/2$ $\endgroup$ Mar 11, 2012 at 18:45
  • $\begingroup$ And isn't $\frac{g}{2}t_x^2$ the same as x? $\endgroup$ Mar 11, 2012 at 18:58
  • $\begingroup$ a) Subsitution works just as well; you can substitute $h$ from $h=gt^2/2$ and then cancel $gt^2/2$ on both sides; the result is the same. b) No, $\frac{g}{2}t_x^2$ is not the same as $x$; if it were, the last equation in my answer would simplify to $gtt_x=0$, which is obviously false in general. $\endgroup$
    – joriki
    Mar 11, 2012 at 19:38

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