# Kinematics - Calculating Distance

1. Oct 11, 2013

### Insolite

Question: "A man steps from the top of a tall building. He falls freely from rest to the ground, a distance of h. He falls a distance of h/4 in the last 0.800 s of his fall. Neglecting air resistance, calculate the height h of the building.

I've attempted the solution in a number of different ways, each of which give different answers.

Solution:
I have labelled the initial time (i.e. at the instant before he begins to fall) as $t_{0}$, the time at which he has traveled $\frac{3}{4}$ of the total distance h as $t_{1}$ and the time at which he has traveled the full height h as $t_{2}$.

Using the equation $x = x_{0} + v_{0}t + \frac{1}{2}a_{x}t^{2}$, where $x = h$, $t = t_{2}$ and $a_{x} = g$.
$x_{0} = 0$ as no distance has been traveled whilst the man is still at the top of the building
$v_{0}t = 0$ as the initial velocity at the top of the building is zero

Therefore -

$h = h_{0} + v_{0}t_{2} + \frac{1}{2}gt_{2}^{2}$
$h = 0 + 0 + \frac{1}{2}gt_{2}^{2}$
$h = \frac{1}{2}gt_{2}^{2}$

Re-arranging to give -

$t_{2} = \sqrt{\frac{2h}{g}}$

$(t_{2} - t_{1})$ = the time to travel $\frac{1}{4}h = 0.800 s$
$(t_{1})$ = the time to travel $\frac{3}{4}h = \sqrt{\frac{2h}{g}} - 0.800 s$

Substituting the value of $t_{2}$ for $h$ in the latter of these equations gives -

$\frac{3}{4}\sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}} - 0.800 s$
Therefore 0.800 must be equivalent to $\frac{1}{4}\sqrt{\frac{2h}{g}}$
$0.800 = \frac{1}{4}\sqrt{\frac{2h}{g}}$
$3.20 = \sqrt{\frac{2h}{g}}$
$h = \frac{3.20^{2}g}{2}$
$h = 50.2 m$

I believe that my mistake is in assuming that $(t_{1}) = \sqrt{\frac{2h}{g}} - 0.800 s$ is equivalent to $\frac{3}{4}h\sqrt{\frac{2h}{g}}$, as the man's velocity increases with constant acceleration and therefore $\frac{3}{4}$ of $t_{2}$ does not correspond to having traveled $\frac{3}{4}h$.

However, I'm uncertain how else to approach the problem. I have tried other methods involving heavier use of the equations of motion with constant acceleration and have twice calculated h as ≈ 12 m, but this would not be a tall building! (I realise that the actual answer does not always have to accurately reflect the real-world scenario described in the question, but I believe I should use it as guidance in evaluating my answer.)

Any help and/or advice would be greatly appreciated.

Last edited: Oct 11, 2013
2. Oct 11, 2013

### NihalSh

Yes, exactly. Up to the point of this assumption everything was correct. And you even know the reason. solve for $t_{1}$ just like you solved for $t_{2}$ but this time take the height to be $\frac{3}{4}h$. Now, substitute this value and solve. You'll arrive at the answer.

P.S. : $t_{1}=\sqrt{\frac{3h}{2g}}$

3. Oct 11, 2013

### Insolite

Hi, thankyou for your response.

I had approached the problem this way previously, but for some reason or another I discarded it.
I have tried to run it through and have calculated what I think is a reasonable answer -

$t_{2} = \sqrt{\frac{2h}{g}}$
$t_{1} = \sqrt{\frac{3h}{2g}}$ (derived from the equation of motion used previously using $\frac{3}{4}h$ as $x$)
$(t_{2} - t_{1}) = 0.800 s$
$t_{2} - (t_{2} - t_{1}) = t_{1}$
$\sqrt{\frac{2h}{g}} - 0.800 s = \sqrt{\frac{3h}{2g}}$
${\left(\sqrt{\frac{2h}{g}} - 0.800 s\right)}^2 = {\left(\sqrt{\frac{3h}{2g}}\right)}^2$
$\frac{2h}{g} - 1.6\sqrt{\frac{2h}{g}} + 0.64 = \frac{3h}{2g}$
$\frac{h}{2g} - 1.6\sqrt{\frac{2h}{g}} + 0.64 = 0$

Solving the quadratic equation using the following substitutions -

$u = \sqrt{2h}$ and $u^{2} = 2h$
$\frac{1}{4g}u^{2} - \frac{1.6}{\sqrt{g}}u + 0.64 = 0$

Using the quadratic formula -
$x = -\frac{b\pm\sqrt{b^{2}-4ac}}{2a}$

$u \approx 18.7026$
$u \approx 1.3429$
$u = \pm\sqrt{2h}$
$h = 175 m$

I hope I haven't made any errors, please correct me where appropriate. I realise that I should try to keep the value of u as an exact number for use in further calculation, but I kept making mistakes when trying to input values in my calculator and found it easiest to right down the values of a, b and c to high precision for use in the formula, as the final answer is required to only three significant figures.

Thanks again for your input

4. Oct 12, 2013

### NihalSh

Yes, its correct. And you have managed to separate out the real solution from the unreal one.

P.S. : Separating the roots is done by checking the values in the original equation involving $t_{1}$ and $t_{2}$