Chapter 3, Problem 11Q

Solution

To Determine:

Whether an arrow should be aimed directly at the target, the way the angle depends on the distance to the target.

Solution:

The arrow should not be aimed directly at the target.

Explanation:

The arrow shot is a parabolic shot and it is necessary to consider the distance of the target to choose the best angle and velocity to fire.

Formula used:

A parabolic shot in two dimensions has the next two equations that describe the movement of the object, in this case, the arrow:

  $\begin{array}{l}x=x_o+v\cos \theta t\\ y=y_o+v\sin \theta t-\frac{1}{2}g{t}^2\end{array}$

Where $x_o$ and $y_o$ are the initial positions in their respective axes, v is the initial velocity, $\theta$ is the angle with the horizontal which the arrow is shot, t is the time since the shot and g is gravity.

Calculations:

Now, let suppose these are the target coordinates:

  $\begin{array}{l}x=x_{\text{target}}\\ y=y_{\text{target}}\end{array}$

If we want our arrow to hit the target, we need to select an angle and a velocity so the equations are equal to the target coordinates.

  $\begin{array}{l}{x}_{\text{target}}=x_o+v\cos \theta t\\ y_{\text{target}}=y_o+v\sin \theta t-\frac{1}{2}g{t}^2\end{array}$

If we shoot the arrow directly to the target, the angle would be 0°, if we substitute this value in equations:

  $\begin{array}{l}{x}_{\text{target}}=x_o+v\cos \left(0\right)t\\ \Rightarrow x_{\text{target}}=x_o+vt\\ \Rightarrow t=\frac{x_{\text{target}}-x_o}{v}\\ y_{\text{target}}=y_o+v\sin \left(0\right)t-\frac{1}{2}g{t}^2\\ \Rightarrow y_{\text{target}}=y_o-\frac{1}{2}g{t}^2\\ y_{\text{target}}=y_o-\frac{1}{2}g{\left(\frac{x_{\text{target}}-x_o}{v}\right)}^2\end{array}$

Where $x_{\text{target}}-x_o$ is the distance between the bow and the target:

  $x_{\text{target}}-x_o=d$

Solving for v to know the velocity necessary to reach the target:

  $v=d\sqrt{\frac{g}{2\left(y_{\text{target}}-y_o\right)}}$

As we can see, the initial height must be higher than the targets. If both heights are equal to the argument of the square root would tend to infinity, or if the target’s height is higher, the velocity would have imaginary terms, and this is not possible.

Equation of projectile trajectory is given by:

  $y-y_o=d\tan \theta -\frac{1}{2}\frac{g{d}^2}{v^2{\cos }^2\theta }$

The angle required to hit the target is:

  $\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$

This can be calculated if the height and distance of the target are measured.

Conclusions:

As we can see, we need to consider the distance to the target to select the appropriate angle.