Chapter 7.2, Problem 81AYU

Artillery A projectile fired into the first quadrant from the origin of a coordinate system will pass through the point $\left(x,y\right)$ at time $t$ according to the relationship $\cot \theta =\frac{2x}{2y+g{t}^2}$ where $\theta =$ the angle of elevation of the launcher and $g=$ the acceleration due to gravity $=32.2feet/secon{d}^2$. An artilleryman is firing at an enemy bunker located 2450 feet up the side of a hill that is 6175 feet away. He fires a round, and exactly $2.27$ seconds later he scores a direct hit.

(a) What angle of elevation did he use?

(b) If the angle of elevation is also given by $\sec \theta =\frac{v_{0}t}{x}$, where $v_0$ is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.

To determine

a. The angle of elevation used by the artillery man to hit the projectile on the enemy bunker.

Answer to Problem 81AYU

Solution:

The angle of elevation used by the artillery man to hit the projectile on the enemy bunker.

$=\theta ={\text{ cot}}^{-1}\left(2x/\left(2y+g{t}^2\right)\right)=22.31°$

Explanation of Solution

Given:

The height of the enemy bunker $y=2450\text{ feet}$.

Horizontal location of the enemy bunker from the projectile artillery man $=x=6175\text{ feet}$.

Time taken for the projectile to hit the enemy bunker $t=2.27\text{ seconds}$.

acceleration due to gravity $g=32.2\text{ feet}/{\text{second}}^2$.

Formula used:

$\text{cot}\theta =2x/\left(2y+g{t}^2\right)$

$\Theta =$ angle of elevation of the launcher.

$x=$ Horizontal location of the enemy bunker from the projectile artillery man (Launcher).

$y=$ The height of the enemy bunker.

$t=$ Time taken for the projectile to hit the enemy bunker.

$g=$ acceleration due to gravity.

$\Theta ={\text{ cot}}^{-1}\left(2x/\left(2y+g{t}^2\right)\right)$

$\Theta =\text{ cot}-1\left(2\text{ x }6175/\left(2\text{ x }2450+32.2 \text{x} {2.27}^2\right)\right)$

$\Theta ={\text{ cot}}^{-1}\left(12350/\left(4900+165.92\right)\right)$

$\Theta ={\text{ cot}}^{-1}\left(12350/5065.92\right)={\text{ cot}}^{-1}\left(x/y\right)$

$X=12350,y=5065.92$

Calculation:

To find $\theta$, we need to find $\text{cos}\theta$ because $y={\text{ cos}}^{-1}x$ has the same range as $y={\text{ cot}}^{-1}x$ except where undefined.

Equation of the circle $=x^2+y^2 =R^2$.

Where $R=\text{ radius of the circle}$.

$R=\surd \left(x^2+y^2\right)=\surd \left({12350}^2+{5065.92}^2\right)=\surd \left(152522500+25663545.44\right)=\surd 178186045.44=13348.6$

Both $x$ and $y$ are positive, therefore $\theta$ lies in the I quadrant.

Using the formula,

$\text{cos}\theta =x/R$

$\text{cos}\theta =12350/13348.6,0<\theta <\pi /2\left(\text{I quadrant}\right)$

${\text{cot}}^{-1}\left(12350/5065.92\right)=\theta ={\text{ cos}}^{-1}\left(12350/13348.6\right)$

${\text{cos}}^{-1}\left(0.9251\right)\approx 22.31°={0.3893}^c$

To determine

b.The muzzle velocity of the artillery piece.

Answer to Problem 81AYU

Solution:

The muzzle velocity of the artillery piece $v_o=2940\text{ feet}/\text{second}$.

Explanation of Solution

Given:

Horizontal location of the enemy bunker from the projectile artillery man $=x=6175\text{ feet}$.

Time taken for the projectile to hit the enemy bunker $t=2.27\text{ seconds}$.

From 81 (a),

Angle of elevation $\theta =22.31°$.

Formula used:

$\text{sec}\theta =\left(v_{o}t\right)/x$

Where $v_o=\text{ the muzzle velocity of the artillery piece}$.

$\text{sec}\left(22.31°\right)=\left(v_o\text{ x }2.27\right)/6175$

Calculation:

Using the formula,

$\text{sec}\theta =1/\text{cos}\theta$

$1/\text{cos}\left(22.31°\right)=\left(v_o\text{ x }2.27\right)/6175$

$\text{cos}\left(22.31°\right)=6175/ \left(v_o\text{ x }2.27\right)$

$\left(v_o\text{ x }2.27\right)=6175/\text{cos}\left(22.31°\right)$

$v_o=6175/\left(2.27\text{ x cos}\left(22.31°\right)\right)$

$v_o=2940\text{ feet}/\text{second}$