Chapter 3, Problem 26P

(a)

To determine

The distance between the target and the bullet.

Answer to Problem 26P

0.86 m

Explanation of Solution

Given:

The initial speed of bullet is 180 m/s target is 75 m away.

Formula Used:

The equation of kinematics are,

  $\begin{array}{l}s=ut+\frac{1}{2}a{t}^2\\ v=\frac{s}{t}\end{array}$

Calculation:

  $\begin{array}{l}\text{Initial velocity }u=180\text{ m/s}\\ \text{Acceleration due to gravity }a=10{\text{ m/s}}^2\end{array}$

Substituting the given values,

  $\begin{array}{l}t=\frac{s}{v}\\ t=\frac{75}{180}\\ t=0.416\end{array}$

Substituting the values,

  $\begin{array}{l}s=0\times 0.416+\frac{1}{2}\times 10\times {\left(0.416\right)}^2\\ s=0.86\end{array}$

Conclusion:

Thus, the bullet will miss the target by 0.86 m

(b)

To determine

Angle aimed by the gun to hit target.

Answer to Problem 26P

0.65°

Explanation of Solution

Given:

The initial speed of gun is 180 m/s

And Horizontal range is 75 m.

Formula Used:

  $R=\frac{u^2\sin 2\theta }{g}$

Calculation:

  $\begin{array}{l}\text{Horizontal range, }R=75\text{ m}\\ \text{Initial speed }u=180\text{ m/s}\\ \text{Acceleration due to gravity,}g=10{\text{m/s}}^2\end{array}$

Substituting the value,

  $\begin{array}{l}75={\left(180\right)}^2\frac{\sin 2\theta }{10}\\ \sin 2\theta =\frac{75\times 10}{{\left(180\right)}^2}\\ \sin 2\theta =0.023\\ \theta =0.65°\end{array}$

Conclusion:

Thus, the gun should be aimed to hit the target is 0.65°.