# Projectile motion 'monkey hunting' simulation

#### Quiz

Q) A hunter tries to shoot a rare monkey hanging from a tree with a dart gun. The hunter has the monkey in his sights. But the monkey notices the hunter and drops from the branch exactly when he hears the hunter blow. Will the hunter hit the monkey or miss the target?

A) The dart hit the monkey.
B) The dart miss above the monkey.
C) The dart miss below the monkey.

Monkey and the dart fell at the same rate, under the same influence of gravity. So the dart will hit the monkey.

#### Projectile motion - monkey hunting simulation

When the next simulation is not visible, please refer to the following link.

A famous problem involving projectile motion is the “monkey and hunter problem” (Fig. 1). A hunter spots a monkey hanging from a tree branch, aims his rifle directly at the monkey, and fires. The monkey, hearing the shot, lets go of the branch at the same instant the hunter fires the rifle, hoping to escape by falling to the ground. Will the monkey escape? The unexpected answer is “no”: the bullet will always hit the monkey anyway, regardless of the angle of the rifle, the speed of the bullet, or the distance to the monkey, as long as the monkey is in range.

Fig 1. The “monkey and hunter” problem.

To show that this is so, let’s first define a coordinate system. Let the origin be at the end of the rifle, with the x axis pointing horizontally to the right, and $\ y$ pointing vertically upward. Let $\ D$ be the horizontal distance of the monkey from the origin, and $\ H$ be the initial height of the monkey (Fig. 1).
Now we’ll show that at $\ x=D$, the monkey and the bullet will both be at the same height $\ y$. Let the muzzle velocity of the rifle be $\ v_0$, and let the angle of fire of the rifle from the horizontal be . Let’s begin by finding the time $\ t$ needed for the bullet to travel the horizontal distance $\ D$ from the rifle. Since the horizontal component of the velocity is a constant $\ v0 cos\theta$, the $\ x$ coordinate of the bullet at time $\ t$ is

$\ x_{b} (t) = (v_0 cos\theta)t$ -- (1)

Setting $\ x_{b} (t) =D$ and solving for the time $\ t$, we find (calling this time $\ t_f$ )

$\ t_f = \frac{D}{v_0 cos \theta}$ -- (2)

Next, let’s find the $\ y$ coordinate of the bullet at this time $\ t_f$ .

By
$\ y_{b}(t) = -\frac{1}{2}gt^2 + (v_0 sin \theta)t$ -- (3)

Substituting $\ t = t_{f} = D/(v_0 cos\theta)$, we have

$\ y_b = -\frac{1}{2}g(\frac{D}{v_0cos\theta })^2 + (v_0 sin\theta) (\frac{D}{v_0cos\theta})$ -- (4)
= $\ -\frac{gD^2}{2v_0^2 cos^2 \theta} + D tan \theta$ -- (5)

But from trigonometry, $\ tan\theta = H/D$; making this substitution in the second term on the right, we have

$\ y_{b} = H - \frac{gD^2}{2v_0^2 cos^2 \theta}$ (bullet) -- (6)

Finally, let’s find the $\ y$ coordinate of the monkey at time $\ t_f$ . The monkey falls in one dimension; its $\ y$ coordinate at time $\ t$ is (using Eq. $\ y(t) = \frac{1}{2}at^2 + v_{0}t + y_{0}$  with $\ a= - g, v_{0}=0$ and $\ y_{0} = H$):

$\ y_{m} (t) = H - \frac{1}{2}gt^2$ -- (7)

Now substituting $\ t = t_{f} = D/(v_0 cos\theta)$,

$\ y_{m} = H - \frac{1}{2}g(\frac{D}{v_0cos\theta })^2$ -- (8)

$\ y_{m} = H - \frac{gD^2}{2v_0^2 cos^2 \theta}$ (monkey) -- (9)

Comparing Eqs. (9.53) and (9.56), you can see that the monkey and bullet will have the same $\ y$ coordinate when $\ x=D$, so the monkey will always get hit, regardless of the values of $\ D, H, v_{0}$, or $\ \theta.$

Essentially what’s happening here is that the monkey and bullet are both accelerated by the same amount, $\ g= 9.8 m/s^2$, so for a given amount of time, the monkey will fall the same distance as the bullet falls from the straight-line path it would take if there were no gravity. Therefore, the bullet always hits the monkey.