Chapter 8.7, Problem 23E

a.

To determine

To find: the initial velocity of a shell fired from ground level.

Answer to Problem 23E

  $140.7\text{ feet per second}$

Explanation of Solution

Given information:

Vertical component $\left(y\right)=300\text{ feet}$

Time $\left(t\right)=4.8\text{ seconds}$

  $\theta =82°$ .

Calculation:

The initial velocity $\left|\overrightarrow{v}\right|$ can be evaluated as follows:

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\\ 300=\left(4.8\right)\left|\overrightarrow{v}\right|\sin 82°-\frac{1}{2}\left(32\right){\left(4.8\right)}^2\left[\text{Put the value of }y,g,t\text{ and }\theta \right]\\ 300=\left(4.8\right)\left|\overrightarrow{v}\right|\sin 82°-368.64\\ 368.64+300=\left(4.8\right)\left|\overrightarrow{v}\right|\sin 82°\\ \left|\overrightarrow{v}\right|=\frac{668.64}{4.8\sin 82°}\\ =140.7\text{ feet per second}\end{array}$

Thus the initial velocity is $\text{140}\text{.7 feet per second}$ .

b.

To determine

To find: the distance between the barrier and the explosion shell.

Answer to Problem 23E

131.3 yards

Explanation of Solution

Given information:

Range of barrier $\left(r\right)$ = 100 yards

Time $\left(t\right)=4.8\text{ seconds}$

  $\theta =82°$ .

  $\left|\overrightarrow{v}\right|=140.7\text{ feet per second}$

Calculation:

The distance between the barrier and explosion shell $\left(x\right)$ will be evaluated as follows:

  $\begin{array}{c}x=\left|\overrightarrow{v}\right|\cos \theta +r\\ =\left(4.8\right)\left(140.7\text{ ft/s}\right)\cos 82°+100\text{ yards }\left[\text{Put the value of }t,r,\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ \text{=}\frac{1}{3}\left(4.8\right)\left(140.7\right)\cos 82°+100\text{ yards }\left[\because 1\text{foot}=\frac{1}{3}\text{yards}\right]\\ =131.3\text{ yards}\end{array}$

Thus the distance is 131.3 yards.