Chapter 8, Problem 3P

Path of a Rocket Suppose that a rocket is fired at an angle of 5° from the vertical with an initial speed of 1000 ft/s.

1. (a) Find the length of time the rocket is in the air.
2. (b) Find the greatest height it reaches.
3. (c) Find the horizontal distance it has traveled when it hits the ground.
4. (d) Graph the rocket’s path.

To determine

To find: length of time of the rocket in the air.

Answer to Problem 3P

The length of time of rocket in air is $203.26\text{sec}$.

Explanation of Solution

Given:

Initial velocity of the rocket, $v_o=1000\text{ft/s}$.

Projection angle of the rocket with vertical, $\theta =5°$.

Calculation:

Calculate the horizontal and vertical components of velocity of the rocket.

The rocket is projected with an angle of $5°$ with vertical so, the projection of the rocket with horizontal is,

$\begin{array}{c}{\theta }_H=90°-{\theta }_V\\ =90-5\\ =85°\end{array}$

The horizontal component of the velocity is

$\begin{array}{c}{v}_x=v_o\cos \theta \\ =v_o\cos 85°\end{array}$

The vertical component of the velocity is,

$\begin{array}{c}{v}_y=v_o\sin \theta \\ =v_o\sin 85°\end{array}$

The time of flight of the projectile is given by the formula.

$t=\frac{2v_o\sin \theta }{g}$

Where,

• $v_o$ is the initial velocity of the projectile
• $\theta$ is the projection angle with horizontal.
• $g$ is the acceleration due to gravity.

Substitute $1000\text{ft/sec}$ for $v_o$, $85°$ for $\theta$ and $9.8{\text{m/sec}}^2$ in the above formula.

$\begin{array}{c}t=\frac{2\times 1000\times \sin 85}{9.8}\\ =\frac{0.996\times 2000}{9.8}\\ =203.26\text{sec}\end{array}$

Hence the length of time of rocket in air is $203.26\text{sec}$.

To determine

The maximum height attained by the rocket.

Answer to Problem 3P

The maximum height attained by the rocket is $50612.24\text{ft}$.

Explanation of Solution

The maximum height of a projectile is given by the formula

$y_{\max }=\frac{v_o^2{\sin }^2\theta }{2g}$

Substitute $1000\text{ft/sec}$ for $v_o$, $85°$ for $\theta$ and $9.8{\text{m/sec}}^2$ in the above formula.

$\begin{array}{l}{y}_{\max }=\frac{{\left(1000\right)}^2{\sin }^2\left(85°\right)}{2\times 9.8}\\ =\frac{1000000\times 0.992}{19.6}\\ =50612.24\text{ft}\end{array}$

Hence the maximum height attained by the rocket is $50612.24\text{ft}$.

To determine

The horizontal distance covered by the rocket.

Answer to Problem 3P

The horizontal distance covered by the rocket is $17719.20\text{ft}$.

Explanation of Solution

The horizontal distance covered by the projectile is given as,

$x=\frac{v_o^2\sin 2\theta }{g}$

Substitute $1000\text{ft/sec}$ for $v_o$, $85°$ for $\theta$ and $9.8{\text{m/sec}}^2$ in the above formula.

$\begin{array}{c}x=\frac{{\left(1000\right)}^2\sin \left(2\times 85\right)}{9.8}\\ =\frac{173648.17}{9.8}\\ =17719.20\text{ft}\end{array}$

Hence the horizontal distance covered by the rocket is $17719.20\text{ft}$.

To determine

To sketch: The graph of the path of rocket.

Explanation of Solution

The parametric equation of the path of rocket is given as

$y=x\tan \theta -\frac{1}{2v_o\cos \theta }{x}^2$

Substitute, $1000\text{ft/sec}$ for $v_o$ and $85°$ for $\theta$ in the above formula.

$\begin{array}{l}y=x\tan 85°-\frac{1}{2\times 1000\times \cos 85°}{x}^2\\ y=11.43x-\frac{x^2}{174.31}\end{array}$

Draw the graph of the above function,

Figure (1)

Hence, Figure (1) represents the path of rocket.