Chapter 8, Problem 4P

Firing a Missile The initial speed of a missile is 330 m/s.

1. (a) At what angle should the missile be fired so that it hits a target 10 km away? (You should find that there are two possible angles.) Graph the missile paths for both angles.
2. (b) For which angle is the target hit sooner?

To determine

The angle of the missile to be fired that hits a target 10 km away and the graph of the missile paths for both angles.

Answer to Problem 4P

The angle of the missile to be fired that hits a target 10 km away is $32.08°,57.93°$.

Explanation of Solution

Given:

The initial speed of a missile is 330 m/s and the distance of the missile from the target is 10 km.

Formula used:

The formula of the parametric equations for the path of the particle with gravity and an angle $\theta$ is,

$x=\left({\nu }_0\cos \theta \right)t$ (1)

And

$y=\left({\nu }_0\sin \theta \right)t-\frac{1}{2}g{t}^2$ (2)

Calculation:

Consider the initial speed be ${\nu }_0$ and gravity $g=9.8{\text{m/s}}^{\text{2}}$.

The given distance of the missile from the target is 10 km.

Convert km into m.

$\begin{array}{c}x=10\times 1000\\ =10000\text{m}\end{array}$

Substitute 330 m/s for ${\nu }_0$, 10,000 m for x in equation (1),

$\begin{array}{c}10000=\left(330\cos \theta \right)t\\ t=\frac{10000}{330\cos \theta }\end{array}$

Substitute 330 m/s for ${\nu }_0$, 0 for y, $9.8{\text{m/s}}^{\text{2}}$ for g, $\frac{10000}{330\cos \theta }$ for t in equation (2),

$\begin{array}{c}0=\left(330\sin \theta \right)\left(\frac{10000}{330\cos \theta }\right)-\frac{1}{2}\left(9.8\right){\left(\frac{10000}{330\cos \theta }\right)}^2\\ =330\sin \theta \left(\frac{10000}{330\cos \theta }\right)-4.9\left(\frac{{\left(10000\right)}^2}{{\left(330\right)}^2{\cos }^2\theta }\right)\\ =330\sin \theta \left(\frac{10000}{330\cos \theta }\right)-4.9\left(\frac{{\left(10000\right)}^2}{108900\times {\cos }^2\theta }\right)\end{array}$

Multiply both sides by ${\cos }^2\theta$,

$\begin{array}{c}0=\left[330\sin \theta \left(\frac{10000}{330\cos \theta }\right)-4.9\left(\frac{{\left(10000\right)}^2}{108900\times {\cos }^2\theta }\right)\right]{\cos }^2\theta \\ =10000\sin \theta \cos \theta -4.9\left(\frac{{\left(10000\right)}^2}{108900}\right)\\ =10000\left(\frac{1}{2}\sin 2\theta \right)-4499.54\\ =5000\sin 2\theta -4499.54\end{array}$

Calculate $\sin 2\theta$ from the above equation.

$\begin{array}{c}\sin 2\theta =\frac{4499.5}{5000}\\ 2\theta ={\sin }^{-1}\left(\frac{4499.5}{5000}\right)\\ 2\theta =64.15°,115.85°\\ \theta =32.08°,57.93°\end{array}$

Thus, the angle of the missile to be fired that hits a target 10 km away is $32.08°,57.93°$.

Sketch the graph of both paths of the missile.

Figure (1)

Thus, Figure (1) shows the graph of both the paths of the missile.

To determine

The angle at which the target hits sooner.

Answer to Problem 4P

The smaller angle at which the target hits sooner is $32.08°$.

Explanation of Solution

Given:

The initial speed of a missile is 330 m/s and the distance of the missile from the target is 10 km.

Calculation:

The angles of the missile to be fired that hits a target 10 km away is $\theta =32.08°,57.93°$ from part (a).

The smaller angle is the angle at which target hits sooner.

Thus, the smaller angle at which the target hits sooner is $32.08°$.