×
Classical Mechanics

# Projectile motion with Robin Hood

Robin Hood was a beguiling casanova who sometimes lived in a tree, but he was also a master archer who used his bow and arrow to challenge nemeses and rescue his friends from danger.

In this quiz, we'll model Robin Hood's shots using projectile motion and see how, precisely, he pulls it off.

Near the surface of Earth, the gravitational field is roughly constant and is equal to $$g$$. This causes all objects to accelerate downward with $$a_y = -g$$ toward Earth, including Robin Hood's arrows.

Though not equal to the measured value, we'll take $$g$$ to be $$\SI[per-mode=symbol]{10}{\meter\per\second\squared}$$ throughout this quiz for ease of calculation.

Robin Hood is known for stealing things from the rich to give them to the poor. He finds himself surrounded by some angry noblemen who demand he return their gold chest.

Outnumbered, he takes an arrow from his quiver and fires it straight into the air, scattering the noblemen who are afraid it will hit them on the way down.

How much time (in seconds) does he have before the arrow hits the ground?

Details and Assumptions:

• The arrow's launch velocity is $$v = \SI[per-mode=symbol]{120}{\meter\per\second}$$.
• Assume that Robin Hood shoots the arrow from ground level.

As we alluded at the end of the Derive Quiz, our results work in multiple dimensions. Recall, the vector position obeys $\mathbf{r}(t) = \mathbf{r}(0) + \mathbf{v}(0)t + \frac12 \mathbf{a}_0 t^2.$ Motion along each dimension is independent so that \begin{align} r_x(t) &= r_x(0) + v_x(0)t + \frac12 a_x t^2 \\ r_y(t) &= r_y(0) + v_y(0)t + \frac12 a_y t^2. \end{align} To visualize the relationship between a vector and its components, consider the velocity of the arrow below.

In the vector view, our arrow travels with speed $$\lvert\mathbf{v}\rvert$$ at an angle $$\theta$$ to the horizontal. Broken into components, it has speed $$\lvert \mathbf{v} \rvert \cos\theta$$ in the $$x$$ direction and speed $$\lvert \mathbf{v} \rvert \sin\theta$$ in the $$y$$ direction.

The independent relations for $$r_x$$ and $$r_y$$ are all there is to projectile motion.

Fresh off his daring escape, Robin Hood found a band of Merry Men. They agree to join him if he can shoot an arrow further than any of them can.

How far (in $$\si{\meter}$$) will his arrow go?

Details and Assumptions

• Ignore Robin Hood's height, e.g. he shoots the arrow while lying down.
• The arrows releases at $$v = \SI[per-mode=symbol]{120}{\meter\per\second}$$ at an angle of $$60^\circ.$$

If Robin Hood wants to maximize the range of his arrow, at what angle $$\theta$$ (in degrees) should he aim his bow?

Details and Assumptions

• Ignore Robin Hood's height, e.g. he shoots the arrow while lying down.

The Merry Men can't believe how far he shot the arrow. They insist that he one-up himself to prove he didn't cheat, so they place an apple high on a $$\SI{40}{\meter}$$ wall that's $$\SI{400}{\meter}$$ away.

To test his restraint, they force him to angle his bow at $$\theta =60^\circ$$.

Note: figure not drawn to scale

How fast (in $$\si[per-mode=symbol]{\meter\per\second}$$) should he shoot the arrow if he wants to hit the apple?

Details and Assumptions

• Ignore Robin Hood's height, e.g. he shoots the arrow while lying down.

In the last problem, we solved for the case where Robin Hood's angle was constrained. Suppose we constrained him to fire at full speed, $$v=\SI[per-mode=symbol]{120}{\meter\per\second}$$.

At what angle should he release the arrow?

Hint: You may not be able to do this with analysis alone, so consider resorting to a plot. As this problem is quite tricky, we'll solve it together on the next pane.

Once again, we start with \begin{align} \ell &= v\cos\theta T \\ H &= v\sin\theta T - \frac12 gT^2. \end{align} We can use the first equation to eliminate $$T$$ in the second: \begin{align} H &= \ell\tan\theta - \frac12 g\frac{\ell^2}{v^2\cos^2\theta} \\\\ 2v^2\cos^2\theta H &= 2\ell\tan\theta v^2\cos^2\theta - g\ell^2 \\\\ 2v^2\cos^2\theta\left(\ell\tan\theta - H\right) &= g\ell^2. \end{align} This equation is a pain to solve analytically for $$\theta$$ and is easier to resolve by numerical means. We can plot the functions on either side and look for an intercept.

We find that the plots intersect at roughly $$\theta \approx 13.75^\circ$$ and $$81.36^\circ$$.

In the last problem we saw that setting a launch speed does not pick out a specific launch angle, despite the fact that setting a launch angle picks out a specific launch velocity.

Why is this? The answer is clear from a drawing.

If an arrow has enough speed to hit the apple on the way up (before its peak) then it has enough speed to hit it on the way down. Therefore, any speed $$v$$ will have two solutions, $$\theta_+$$ and $$\theta_-,$$ corresponding to a hit before and after the trajectory's peak. At the minimal velocity $$v_\textrm{min}$$ that's still capable of hitting the apple, the two angles coincide.

Here we argued for the existence of two solutions informally.

• Can you take it a step further and prove this result?
• Can you find $$v_\textrm{min}$$?

Having gained their allegiance, the Merry Men agree to help Robin Hood take down a corrupt noblewoman who reigns over the valley with onerous taxes. Each night she sleeps alone on her porch that overlooks the valley, as her guards tell each other jokes nearby.

What is the smallest angle, $$\gamma = \phi + \theta$$ (in degrees), the crew should aim their arrows to hit the noblewoman?

Details and Assumptions:

• Each archer shoots their arrow at full speed, $$v_0 = \SI[per-mode=symbol]{120}{\meter\per\second}$$.
• The bed is located a distance $$s = \SI{400}{\meter}$$ up the hill (measured along the hill) and the hill is pitched at $$\theta=60^\circ$$.
• Ignore the height of the archers, e.g. each person shoots their arrow while lying down.

In helping Robin Hood win over the Merry Men and take down the punitive nobles, we have solved some of the classic hard problems of projectile motion.

Though we considered the projectile motion of the point particle (as an idealization of our arrow), we used the same basic building blocks as the techniques used to put satellites into orbit, dock the space station, and chart the course of passenger planes.

By breaking the problem down into the horizontal and vertical dimensions, we were able to treat the arrow's complete motion as two independent, one-dimensional problems, a process known as vector decomposition. As you will see in the next chapter, this step reflects a deep principle of mechanics embedded in Newtons laws of motion—that motion in a given dimension is controlled solely by forces along that dimension.

×