Chapter 3.5, Problem 37AYU

Artillery A projectile Fired from the point $\left(0,0\right)$ at an angle to the positive x-axis has a trajectory given by

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

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 (m/sec2)

$c>0$ 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?

Source: www. answers, com

To determine

To find:

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?

Answer to Problem 37AYU

Solution:

a. $\left\{c|0.112<c<81.912\right\}$

b. It is possible to hit a target 75 km away if $c=0.651,1.536$.

Explanation of Solution

Given:

A projectile fired from the point $\left(0,0\right)$ at an angle to the positive $x\text{-axis}$ has a trajectory given by

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

$x$ = 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 (m/sec)

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

Calculation:

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

$v=897$

$g=9.81$

$x=2000$

$y=200$

$200=c2000-\left(1+c^2\right)\left(\frac{9.81}{2}\right){\left(\frac{2000}{897}\right)}^2$

$200=c2000-\left(1+c^2\right)\left(4.905\right)\left(4.971\right)$

$200=c2000-\left(1+c^2\right)\left(24.383\right)$

$200=c2000-\left(24.383+24.383c^2\right)$

$0=-24.383c^2+2000c-224.383$

Using the formula

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here

$a=-24.383, b=2000,c=-224.383$

Therefore

$-2000\pm \frac{1994.521}{-48.766}$

That is

$-2000+\frac{1994.521}{-48.766}$ and $-2000-\frac{1994.521}{-48.766}$

The roots are

$0.112$ and $81.912$

Therefore $\left\{c|0.112<c<81.912\right\}$

b. It is possible to hit a target 75 km away so we use the distance formula and simplify the following solution

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

$200=c\left(75000\right)-\left(1+c^2\right)\left(\frac{9.81}{2}\right){\left(\frac{75000}{897}\right)}^2$

$c^2-1.6011c+0.0999936=0$

Then we use the roots equation we get

$c=0.651$ and $c=1.536$

It is possible to hit a target 75 km away if $c=0.651, 1.536$.