Question

An enemy ship is on the east side of a mountain island as shown inFigure$\mathit{P}\mathbf{4}\mathbf{.89}$. Theenemy ship has maneuvered to within $\mathbf{2500}\mathbf{m}$of the$\mathbf{1800}\mathbf{-}\mathbf{m}$-high mountain peak and can shoot projectiles with an initial speed of$\mathbf{250}\mathbf{m}\mathbf{/}\mathbf{s}$. If the western shoreline ishorizontally$\mathbf{300}\mathbf{m}$from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

Short answer

$<265.08\text{m}$The safe distance is $>3476\text{m}$and from the shore.

Step-by-step solution

Definition of Range of projectile.

• The horizontal displacement of the projectile determines its maximum or minimum range. Gravity most effective acts vertically, consequently there may be no acceleration on this direction indicates the variety line.
• The variety of the projectile, like its time of flight and most height, is a feature of its preliminary speed.
• $\mathit{R}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{\theta }}{\mathbf{g}}$
  

Known and the given values.

• As we can see from the figure, the angle${\theta }_{\text{H}}$gives the enemy the minimum range satisfying the condition of passing the mountaintop.
• The angle${\theta }_{\text{L}}$gives the enemy the maximum range satisfying the condition of passing the mountaintop.
• So our ship will be safe between the western shore and the point of the minimum range, and beyond the point of the maximum range.

• We have the initial velocity of the projectile,$v_i=250\text{m}/\text{s}$the mountaintop has coordinates$(x_t=2500\text{m},y_t=1800\text{m})$ and the horizontal distance between the peak and the western shoreline is $300\text{m}.$

Determine the kinematic equation for vertical and horizontal motion. 

Let$\theta$be an arbitrary angle that satisfying the condition of passing the mountaintop. So the components of the initial velocity are:

$v_x=v_i\cos \theta =250\cos \theta$

$v_{yi}=v_i\sin \theta =250\sin \theta$

For the horizontal motion, we substitute the given values into the kinematic equation in order to get the time it takes to reach the mountaintop as follows:

$x_f=x_i+v_{x}t+\frac{1}{2}{a}_x{t}^2$

$2500=0+\left(250\cos \theta \right)\times t+0$

$t=\frac{10}{\cos \theta }$

For the vertical motion, we substitute the given values into the kinematic equation so we get:

$y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2$

$1800=0+\left(250\sin \theta \right)\times t+\frac{1}{2}(-9.8)t^2$

Substituting with the value of we have calculated above, we get:

$1800=\left(250\sin \theta \right)\times \left(\frac{10}{\cos \theta }\right)-4.9{\left(\frac{10}{\cos \theta }\right)}^2$

$1800=2500\tan \theta -490{\sec }^2\theta$

But we know that${\sec }^2\theta =1+{\tan }^2\theta$, so we have:

$\begin{array}{l}1800=2500\tan \theta -490-490{\tan }^2\theta \\ 490{\tan }^2\theta -2500\tan \theta +2290=0\end{array}$

By using the Quadratic equation.

This is a quadratic equation which can be solved using the formula

$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}$

Where $a=490,b=-2500$and $c=2290,$so we get:

$\tan \theta =\frac{2500\pm \sqrt{{(-2500)}^2-(4\times 490\times 2290)}}{2\times 490}$

$=1.2\&3.9$

Now, we have two solutions $\tan \theta =1.2$which gives us ${\theta }_{\text{L}}$as follows:

${\theta }_L={\tan }^{1}1.2={50.11}^{°}$

and $\tan \theta =3.9$gives us ${\theta }_{\text{H}}$as follows:

${\theta }_{\text{H}}={\tan }^{-1}3.9={75.64}^{°}$

Determine the minimum and maximum range of projectile.

We know that the range of a projectile is given by equation:

$\mathit{R}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{\theta }}{\mathbf{g}}$

By substituting${\theta }_L$we get the maximum range:

$\begin{array}{l}{R}_{\text{M}}=\frac{{\left(250\right)}^2\sin (2\times 50.11)}{9.8}\\ R_M=6276\text{m}\end{array}$

So our ship will be safe at:

$\begin{array}{l}d>6276-(2500+300)\\ \therefore d>3476\text{m\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

From the shoreline.

And by substituting${\theta }_{\text{H}}$we get the minimum range:

$\begin{array}{l}{R}_{\text{m}}=\frac{{\left(250\right)}^2\sin (2\times 75.64)}{9.8}\\ R_m=3065.08\text{m}\end{array}$

So our ship will be safe at :

$d<3065.08-(2500+300)$

$\therefore d<265.08\text{m}$from the shoreline

Therefore,the safe distance is $>3476\text{m}$and $<265.08\text{m}$from the shore.