Chapter 9.8, Problem 14P

To determine

To find: the distance travelled by the rock from the place where it was thrown.

Answer to Problem 14P

The distance travelled by the rock from the place where it was thrown is $262.39m$ .

Explanation of Solution

Given:

  $y=300-\frac{x^2}{80}$

Calculation:

The given canyon wall is drawn below:

  

The rock follows a path described approximately by the equation

  $y=300-\frac{x^2}{80}\mathrm{.........}\left(A\right)$

The given figure can be representing in coordinate system as below:

  

The line connecting the points $\left(10,0\right)$ and $\left(510,300\right)$ is:

  $\begin{array}{l}\left(y-0\right)=\frac{300-0}{510-10}\left(x-10\right)\\ y=\frac{300}{500}\left(x-10\right)\\ y=\frac{3}{5}\left(x-10\right)\end{array}$

The intersection point between the above line and the curve representing equating (A) is given as follows:

  $\begin{array}{l}y=300-\frac{x^2}{80}\\ \frac{3}{5}\left(x-10\right)=300-\frac{x^2}{80}\left[\mathrm{Replacing}y=\frac{3}{5}\left(x-10\right)\right]\\ \frac{3}{5}\times 80\left(x-10\right)=80\left[300-\frac{x^2}{80}\right]\left[Multiplyingbothsideby80\right]\\ 3\times 16\left(x-10\right)=24000-x^2\\ 48\left(x-10\right)=24000-x^2\\ 48x-480=24000-x^2\\ x^2+48x-480-24000=0\\ x^2+48x-24480=0\\ x=\frac{-48\pm \sqrt{{\left(48\right)}^2-4\times \left(-24480\right)}}{2}\\ =\frac{-48\pm \sqrt{100224}}{2}\\ =\frac{-48\pm 316.582}{2}\\ =134.291and-182.291\end{array}$

Distance can't be negative.

Therefore,

  $x=134.291$

Now, at $x=134.291$ ,

  $\begin{array}{l}y=\frac{3}{5}\left(x-10\right)\\ =\frac{3}{5}\left(134.291-10\right)\\ =74.5746\end{array}$

Therefore, the point is $\left(134.291,74.5746\right)$ .

The distance of the point $\left(134.291,74.5746\right)$ from the point $\left(0,300\right)$ is given as follows:

  $\begin{array}{l}d=\sqrt{{\left(300-74.5746\right)}^2+{\left(0-134.291\right)}^2}\\ =\sqrt{{\left(225.4254\right)}^2+{\left(-134.291\right)}^2}\\ =\sqrt{50816.61+18034.07}\\ =262.39\end{array}$

Therefore, distance is $262.39m$ .

Conclusion:

Therefore, the distance travelled by the rock from the place where it was thrown is $262.39m$ .