Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

Artillery A projectile fired from the point (0, 0) at an angle

to the positive x-axis has a trajectory given by

$y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2$

where

x = horizontal distance in meters

y = height in meters

v = initial muzzle velocity in meters per second (m/sec)

g = acceleration due to gravity = 9.81 meters per second squared $\left(m/se{c}^2\right)$ 

$c>0$ is a constant determined by the angle of elevation.

A howitzer fires an artillery round with a muzzle velocity of 897 m/sec.

(a) If the round must clear a hill 200 meters high at a distance of 2000 meters in front of the howitzer, what c values are permitted in the trajectory equation?

(b) If the goal in part (a) is to hit a target on the ground 75 kilometers away, is it possible to do so? If so, for what values of c? If not, what is the maximum distance the round will travel?

Step-by-Step Solution

Answer

Part a. The permitted values of c in the trajectory equation are $0.112<c<81.9.$

Part b. Yes, it is possible to hit a target on the ground $75$ km away if the values of c are $0.651 \mathrm{or} 1.536.$

Part (a) Step 1. Given Information

The given trajectory is $y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2.$

The velocity is $897 m/sec.$ We have to find the permitted values of c in the trajectory equation if the round must clear a hill 200 meters high at a distance of 2000 meters in front of the howitzer.

Part (a) Step 2. Finding the values

To find the values substitute all the values in the trajectory equation.

$$\begin{gathered} y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2 \\ 200=c\left(2000\right)-\left(1+c^2\right)\left(\frac{9.81}{2}\right){\left(\frac{2000}{897}\right)}^2 \\ 200=2000c-\left(1+c^2\right)\left(4.905\right)\left(4.97\right) \\ 200=2000c-\left(1+c^2\right)\left(24.37\right) \\ 200=2000c-24.37-24.37c^2 \\ 2000c-24.37-24.37c^2-200=0 \\ 2000c-24.37c^2-224.37=0 \end{gathered}$$

Part (a) Step 3. Solve

We will use the formula of the quadratic equation to find the value of c. By proceeding with the calculation further we get,

$$\begin{gathered} 2000c-24.37c^2-224.37=0 \\ {b}^2-4ac={\left(2000\right)}^2-4\left(-24.37\right)\left(-224.37\right) \\ {b}^2-4ac=4000000-21871.58 \\ {b}^2-4ac=3978128.42 \end{gathered}$$

Let's use the formula

$$\begin{gathered} c=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ c=\frac{-2000\pm \sqrt{3978128.42}}{2\left(-24.37\right)} \\ c=\frac{-2000\pm 1994.52}{-48.74} \\ c=\frac{-2000+1994.52}{-48.74} or \frac{-2000-1994.52}{-48.74} \\ c=0.112 or 81.9 \end{gathered}$$

Part (b) Step 1. Finding the values of c

To find the values substitute all the values in the trajectory equation.

First, convert 75 kilometers to meters so, $75000m.$

$$\begin{gathered} y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2 \\ 200=c\left(75000\right)-\left(1+c^2\right)\left(\frac{9.81}{2}\right){\left(\frac{75000}{897}\right)}^2 \\ 200=c\left(75000\right)-\left(1+c^2\right)\left(4.905\right)\left(6990.97\right) \\ 200=75000c-\left(34290.71\right)\left(1+c^2\right) \\ 75000c-34290.71-34290.71c^2-200=0 \\ 75000c-34290.71c^2-34490.71=0 \end{gathered}$$

Part (b) Step 2. Solve

We will use the formula of the quadratic equation to find the value of c. By proceeding with the calculation further we get,

$$\begin{gathered} 75000c-34290.71c^2-34490.71=0 \\ {b}^2-4ac={\left(75000\right)}^2-4\left(-34290.71\right)\left(-34290.71\right) \\ {b}^2-4ac=5625000000-4703411169 \\ {b}^2-4ac=921588831 \end{gathered}$$

Let's use the formula

$$\begin{gathered} c=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ c=\frac{-75000\pm \sqrt{921588831}}{2\left(-34290.71\right)} \\ c=\frac{-75000\pm 30357.68}{-68581.42} \\ c=\frac{-75000+30357.68}{-68581.42} or \frac{-75000-30357.68}{-68581.42} \\ c=0.651 or 1.536 \end{gathered}$$

So, it is possible to hit a target 75 km away.