Chapter 7, Problem 49RE

(a)

To determine

To find:The projection angle of fire to attain the maximum height of $2000\text{ft}$ .

Answer to Problem 49RE

Theprojectile should be fired at an angle of $63.4°$ approximately.

Explanation of Solution

Given information:

A projectile is fired with a velocity $v_0$ at angle $\theta$ , then the maximum height it reaches is modeled by the function $M\left(\theta \right)=\frac{v_0^2{\sin }^2\theta }{64}$

Calculation:

Substitute $2000$ for $M\left(\theta \right)$ and $400$ for $v_0$ in the above equation to find the angle.

  $\begin{array}{c}2000=\frac{{\left(400\right)}^2{\sin }^2\theta }{64}\\ 128000=160000{\sin }^2\theta \\ {\sin }^2\theta =0.8\\ \sin \theta =\pm 0.89443\end{array}$

Simplify further.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\pm 0.89443\right)\\ \theta =63.4°\text{or}243.43°\end{array}$

For the maximum height, the projectile must be fired between angle 0 and $90°$ .

Therefore, the projectile should be fired at an angle of $63.4°$ approximately.

(b)

To determine

To check: It is possible for the projectile to reach a height of $3000\text{ft}$ or not.

Answer to Problem 49RE

No, it is not possible for projectile to reach a maximum height of $3000\text{ft}$ .

Explanation of Solution

Given information:

A projectile is fired with a velocity $v_0$ at angle $\theta$ , then the maximum height it reaches is modeled by the function $M\left(\theta \right)=\frac{v_0^2{\sin }^2\theta }{64}$

Calculation:

Substitute $3000$ for $M\left(\theta \right)$ and $400$ for $v_0$ in the above equation to find the angle.

  $\begin{array}{c}3000=\frac{{\left(400\right)}^2{\sin }^2\theta }{64}\\ 192000=160000{\sin }^2\theta \\ {\sin }^2\theta =1.2\\ \sin \theta =\pm 1.09544\end{array}$

It is known that the sine function lies between $-1$ and $1$ .

Therefore, the there is no value of $\theta$ for which the projectile reaches a maximum height of $3000\text{ft}$ .

(c)

To determine

To find:The angle $\theta$ for which the projectile will travel highest.

Answer to Problem 49RE

It should be fired at an angle of $90°$ so that it reaches a maximum height.

Explanation of Solution

Given information:

A projectile is fired with a velocity $v_0$ at angle $\theta$ , then the maximum height it reaches is modeled by the function $M\left(\theta \right)=\frac{v_0^2{\sin }^2\theta }{64}$

Calculation:

The projectile will travel the highest when $M\left(\theta \right)$ is maximum and this happens when ${\sin }^2\theta$ is maximum.

The maximum value of ${\sin }^2\theta$ is equal to 1 which happens when,

  $\begin{array}{c}{\sin }^2\theta =1\\ \sin \theta =\pm 1\\ \theta ={\sin }^{-1}\left(\pm 1\right)\\ =90°\text{or}270°\end{array}$

Since projectile cannot be fired below the ground, it should be fired vertically upward so that it reaches a maximum height.

Therefore, it should be fired at an angle of $90°$ so that it reaches a maximum height.