Chapter 6, Problem 49A

(a)

To determine

The maximum of height attained by the arrow.

Answer to Problem 49A

The maximum height attained by the arrow is $31\text{m}$ .

Explanation of Solution

Given:

The arrow is shot at angle $\theta =30°$ with the horizontal.

The velocity of the arrow is $v_{\text{i}}=49\text{m/s}$ .

Acceleration due to gravity of Earth, $g=9.81{\text{m/s}}^2$

Formula used:

The equation of motion are,

  $\begin{array}{l}v=u+at\\ y_{\max }=ut+\frac{1}{2}a{t}^2\end{array}$

Here, $u$ and $v$ are the initial and final velocity of the object, $a$ is the acceleration and $t$ is the time taken by the object.

Calculation:

An archer shots an arrow upwards in the air.

Consider the figure shown below.

  

The time taken for the motion of arrow to maximum height is,

  $v_{\text{y}}=v_{\text{i}}\sin \theta -gt$

The vertical component of the velocity is zero at the maximum height of arrow’s motion.

  $\begin{array}{l}0=v_{\text{i}}\sin \theta -gt\\ t=\left(\frac{v_{\text{i}}\sin \theta }{g}\right)\end{array}$

The maximum height attained by the arrow is,

  $\begin{array}{l}{y}_{\max }=v_{\text{i}}\sin \theta t-\frac{1}{2}g{t}^2\\ y_{\max }=v_{\text{i}}\sin \theta \left(\frac{v_{\text{i}}\sin \theta }{g}\right)-\frac{1}{2}g{\left(\frac{v_{\text{i}}\sin \theta }{g}\right)}^2\\ y_{\max }=v_{\text{i}}\sin \theta \left(\frac{v_{\text{i}}\sin \theta }{g}\right)-\frac{1}{2}g{\left(\frac{v_{\text{i}}\sin \theta }{g}\right)}^2\\ y_{\max }=\frac{{\left(v_{\text{i}}\sin \theta \right)}^2}{g}-\frac{{\left(v_{\text{i}}\sin \theta \right)}^2}{2g}\\ y_{\max }=\frac{{\left(v_{\text{i}}\sin \theta \right)}^2}{2g}\end{array}$

Substitute the values of velocity, acceleration, and angle of inclination.

  $\begin{array}{l}{y}_{\max }=\frac{{\left(49\times \sin 30°\right)}^2}{2\times 9.80}\\ y_{\max }=\frac{{\left(24.5\right)}^2}{2\times 9.80}\\ y_{\max }=31\text{m}\end{array}$

Conclusion:

Therefore, the maximum height attained by the arrow is $31\text{m}$ .

(b)

To determine

The horizontal distance between shooter and the target.

Answer to Problem 49A

The horizontal distance between shooter and the target is $210\text{m}$ .

Explanation of Solution

Given:

The target is at the height from which the arrow is shot.

The arrow is shot at angle $\theta =30°$ with the horizontal.

The velocity of the arrow is $v_{\text{i}}=49\text{m/s}$ .

Formula used:

The equation of motion are,

  $\begin{array}{l}v=u+at\\ y_{\max }=ut+\frac{1}{2}a{t}^2\end{array}$

Here, $u$ and $v$ are the initial and final velocity of the object, $a$ is the acceleration and $t$ is the time taken by the object.

Calculation:

From part (a), the time taken by the arrow to reach maximum height is,

  $t=\left(\frac{v_{\text{i}}\sin \theta }{g}\right)$

Once the arrow will attain maximum height, it will take the same amount of time to reach the target.

The total time taken by the arrow from the shooter to reach target is

  $\begin{array}{c}t\text{'}=2t\\ =2\left(\frac{v_{\text{i}}\sin \theta }{g}\right)\\ =2\left(\frac{49\times \sin 30°}{9.80}\right)\\ =2\times 2.5\\ =5\text{s}\end{array}$

The horizontal distance $\left(d\right)$ between the shooter and the target is

  $\begin{array}{l}d=\left(v_{\text{i}}\cos 30°\right)t^{\prime }\\ d=\left(49\times \cos 30°\right)\times 5\\ d\approx 210\text{m}\end{array}$

Conclusion:

Therefore, the horizontal distance between shooter and the target is $210\text{m}$ .