Extra Credit: There is an interesting monkey down at the zoo. The monkey spends most of its day hanging from a limb of a tree. The zookeeper feeds the monkey by shooting bananas from a banana cannon to the monkey in the tree. This particular monkey has a habit of dropping from the tree the moment that the banana leaves the muzzle of the cannon. The zookeeper is faced with the dilemma of where to aim the banana cannon in order to hit the monkey. If the monkey lets go of the tree the moment that the banana is fired, then where should she aim the banana cannon? Carefully explain your reasoning and any assumptions you are making. Here are some numbers: The monkey is initially at a height of 15m, the banana cannon is 0m from the tree and shoots the bananas with an initial speed of 10m/s.
The Correct Answer and Explanation is:
To solve this problem, let’s analyze the motion of both the monkey and the banana. The key observation here is that both the monkey and the banana are subject to the same gravitational force, and they will fall at the same rate due to gravity, assuming no air resistance.
Assumptions:
- The monkey begins at a height of 15m.
- The banana cannon is 0m horizontally from the tree.
- The banana is shot with an initial speed of 10 m/s.
- The monkey drops from the tree as soon as the banana is fired.
Solution:
The scenario involves both the monkey and the banana falling under gravity once the banana is fired. The horizontal motion of the banana and the vertical motion of both the monkey and the banana are key to solving this.
Step 1: Vertical Motion of the Banana
The banana is shot with an initial speed of 10 m/s. We’ll break this velocity into horizontal and vertical components. Given that the cannon is aimed at an angle θ to hit the monkey, the vertical velocity component of the banana would be:
vy=10sin(θ) m/sv_y = 10 \sin(\theta) \, \text{m/s}vy=10sin(θ)m/s
Since the banana is shot at an angle, the only force acting on it after the shot is gravity (assuming no air resistance). The vertical motion follows this equation: y(t)=y0+vy0t−12gt2y(t) = y_0 + v_{y0} t – \frac{1}{2} g t^2y(t)=y0+vy0t−21gt2
where:
- y(t)y(t)y(t) is the vertical position of the banana at time ttt,
- y0=15y_0 = 15y0=15 m is the initial height of the banana,
- vy0=10sin(θ)v_{y0} = 10 \sin(\theta)vy0=10sin(θ) m/s is the initial vertical velocity of the banana,
- g=9.8g = 9.8g=9.8 m/s² is the acceleration due to gravity.
Step 2: Vertical Motion of the Monkey
The monkey drops from the tree with an initial velocity of 0 m/s, and it falls under the influence of gravity. Its vertical position ym(t)y_m(t)ym(t) is given by: ym(t)=15−12gt2y_m(t) = 15 – \frac{1}{2} g t^2ym(t)=15−21gt2
Step 3: Condition for the Banana Hitting the Monkey
To hit the monkey, the vertical positions of the banana and the monkey must be equal at some time ttt. Therefore, we want: y(t)=ym(t)y(t) = y_m(t)y(t)=ym(t)
Substituting the expressions for y(t)y(t)y(t) and ym(t)y_m(t)ym(t): 15+10sin(θ)t−12gt2=15−12gt215 + 10 \sin(\theta) t – \frac{1}{2} g t^2 = 15 – \frac{1}{2} g t^215+10sin(θ)t−21gt2=15−21gt2
Simplifying: 10sin(θ)t=010 \sin(\theta) t = 010sin(θ)t=0
This means that sin(θ)=0\sin(\theta) = 0sin(θ)=0, so θ=0∘\theta = 0^\circθ=0∘, meaning that the cannon should be aimed horizontally.
Conclusion:
The zookeeper should aim the banana cannon directly at the monkey horizontally. This ensures that the banana and the monkey will both fall at the same rate and meet at the same point in space. Since both the banana and the monkey are subject to the same gravitational acceleration, they will fall toward the ground at the same rate, and the banana will reach the monkey if shot horizontally.
