Chapter 8.7, Problem 21E

a.

To determine

To find: the maximum range of the cannon.

Answer to Problem 21E

323.2 feet

Explanation of Solution

Given information:

Initial speed $\left|\overrightarrow{v}\right|=70\text{ mph}$

  $\begin{array}{l}\theta =35°\\ h=10\text{ feet}\end{array}$

Calculation:

First convert the velocity into feet per second as follows:

  $\begin{array}{c}\left|\overrightarrow{v}\right|=70\text{ mph }\left(\frac{5280\text{ ft}}{1\text{ mile}}\right)\left(\frac{1\text{ hour}}{3600\text{ s}}\right)\\ =\text{102}\text{.67 meter per second}\end{array}$

The value of $y$ will be zero when the ball has landed. Now solve the value of $t$ as shown below.

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2+h\\ 0=t\left(102.67\right)\sin \left(35°\right)-\frac{1}{2}\left(32\right)t^2\text{+10 }\left[\text{Put the value of }y,\left|\overrightarrow{v}\right|,g,h\text{ and }\theta \right]\\ 0=16t^2-t\left(102.67\right)\sin \left(35°\right)-\text{10 }\left[\text{Dividing the whole equation by }-1\right]\\ t=\frac{102.67\sin 35°\pm \sqrt{{\left(102.67\sin 35°\right)}^2-4\left(16\right)\left(-10\right)}}{2\left(16\right)}\\ =3.84\text{ second}\end{array}$

Since the horizontal displacement is achieved with the horizontal speed. So evaluate horizontal component to get the maximum range.

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =3.48\left(102.67\right)\cos 35°\left[\text{Put the value of }t,\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =323.2\text{ feet}\end{array}$

Thus the maximum range is 323.2 feet.

b.

To determine

To find: the distance from the launch point.

Answer to Problem 21E

312.4 feet

Explanation of Solution

Given information:

  $\begin{array}{l}\theta =35°\\ h=10\text{ feet}\\ y=8\text{ feet}\\ \left|\overrightarrow{v}\right|=70\text{ mph}=\text{102}\text{.67 meter per second}\end{array}$

Calculation:

First solve the value of $t$ as shown below.

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2+h\\ 8=t\left(102.67\right)\sin \left(35°\right)-\frac{1}{2}\left(32\right)t^2\text{+10 }\left[\text{Put the value of }y,\left|\overrightarrow{v}\right|,g,h\text{ and }\theta \right]\\ 0=16t^2-t\left(102.67\right)\sin \left(35°\right)-2\\ t=\frac{102.67\sin 35°\pm \sqrt{{\left(102.67\sin 35°\right)}^2-4\left(16\right)\left(-2\right)}}{2\left(16\right)}\\ =3.71\text{ second}\end{array}$

Since the horizontal displacement is achieved with the horizontal speed. So evaluate horizontal component to get the maximum range.

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =3.71\left(102.67\right)\cos 35°\left[\text{Put the value of }t,\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =312.4\text{ feet}\end{array}$

Thus the distance from the launch point is 312.4 feet.

c.

To determine

To find: the time of flight of human cannonball once it is launched

Answer to Problem 21E

3.71 seconds

Explanation of Solution

Given information:

  $\begin{array}{l}\theta =35°\\ h=10\text{ feet}\\ y=8\text{ feet}\\ \left|\overrightarrow{v}\right|=70\text{ mph}=\text{102}\text{.67 meter per second}\end{array}$

Calculation:

The time of flight $t$ will be evaluated as follows:

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2+h\\ 8=t\left(102.67\right)\sin \left(35°\right)-\frac{1}{2}\left(32\right)t^2\text{+10 }\left[\text{Put the value of }y,\left|\overrightarrow{v}\right|,g,h\text{ and }\theta \right]\\ 0=16t^2-t\left(102.67\right)\sin \left(35°\right)-2\\ t=\frac{102.67\sin 35°\pm \sqrt{{\left(102.67\sin 35°\right)}^2-4\left(16\right)\left(-2\right)}}{2\left(16\right)}\\ =3.71\text{ second}\end{array}$

Thus the time of flight is 3.71 seconds.