# Question a307c

Jun 11, 2015

The total time of its fall is equal to ${t}_{\text{total" = (2+sqrt(2))"s}}$.
The total distance covered is equal to $h = 10 \left(3 + 2 \sqrt{2}\right) \text{m}$.

#### Explanation:

I think that you can go about solving this one in others ways as well, but here's the approach I used.

You know that the object covered half of its total path in 1 second. Moreover, it was the last second of its fall. This means that the distance it covered is actually the bottom half. You can write

${\underbrace{\frac{h}{2}}}_{\textcolor{b l u e}{\text{second half of the fall}}} = {v}_{1} t + \frac{1}{2} g \cdot {t}^{2}$, where

$h$ - the total distance of the fall;
${v}_{1}$ - the speed with which it enters the bottom half of the total distance;
$t$ - the time it took for this distance to be covered - in your case, $t$ is equal to 1 second;
$g$ - the gravitational acceleration, which I'll use as ${\text{10 m s}}^{- 2}$

Use the value of $t$ to determine a relationship between $\frac{h}{2}$ and ${v}_{1}$.

$\frac{h}{2} = {v}_{1} \cdot 1 + \frac{1}{2} \cdot 10 \cdot {1}^{2} = {v}_{1} + 5$ $\text{ } \textcolor{b l u e}{\left(1\right)}$

Now focus on the first half of the fall. You know that your object covers the same distance, i.e. $\frac{h}{2}$, in a time I'll label as ${t}_{1}$.

Once again, you can write

${\underbrace{\frac{h}{2}}}_{\textcolor{b l u e}{\text{first half of the fall}}} = {v}_{0} \cdot {t}_{1} + \frac{1}{2} g \cdot {t}_{1}^{2}$

Since the object starts from a stationary position, ${v}_{0} = 0$. This will get you

$\frac{h}{2} = \frac{1}{2} \cdot 10 \cdot {t}_{1}^{2} = 5 \cdot {t}_{1}^{2}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

For the first half of the fall, you can also write this equation

${v}_{1} = {v}_{0} + g \cdot {t}_{1} \iff {v}_{1} = g \cdot {t}_{1}$

Use this identity to replace ${v}_{1}$ in equation $\textcolor{b l u e}{\left(1\right)}$ to get

$\left\{\begin{matrix}\frac{h}{2} = g \cdot {t}_{1} + 5 = 10 {t}_{1} + 5 \\ \frac{h}{2} = 5 \cdot {t}_{1}^{2}\end{matrix}\right.$

Now solve for ${t}_{1}$.

$5 {t}_{1}^{2} = 10 {t}_{1} + 5 \iff {t}_{1}^{2} - 2 {t}_{1} - 1 = 0$

${t}_{1}^{'} = \frac{2 + \sqrt{4 + 4}}{2} = 1 + \sqrt{2}$

and

${t}_{1}^{' '} = \frac{2 - \sqrt{4 + 4}}{2} = 1 - \sqrt{2}$

Since ${t}_{1}^{' '}$ is negative, the only feasible solution is ${t}_{1}^{'} = 1 + \sqrt{2}$.

This means that the total time of the fall was

t_"total" = t_1 + t = 1 + sqrt(2) + 1 = color(green)(2 + sqrt(2)" seconds")

The total distance covered was

$h = \frac{1}{2} \cdot g \cdot {t}_{\text{total}}^{2} = 5 \cdot {\left(2 + \sqrt{2}\right)}^{2}$

h = 5 * (4 + 4sqrt(2) + 2) = 30 + 20sqrt(2) = color(green)(10(3 + 2sqrt(2)" meters")#