Kinematics - Calculating Distance

[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 realize 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.

 

[NihalSh]

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 $\frac{3}{4}h$.

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

 

[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 realize 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

 

[NihalSh]

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 realize 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

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 $t_{2}$