A grasshopper on the bottom of a cubical box has to jump out of the box. If each side of the box is

Q: Why can't we just use the standard projectile motion formula for maximum height?

The standard formula assumes gravity acts perfectly perpendicular to the horizontal ground. Here, the 'ground' (bottom of the box) is tilted, so gravity acts at an angle relative to it. Tilting the frame makes the math much simpler.

Q: Why does the grasshopper jump perpendicular to the bottom of the box?

To escape, it needs to cover the distance $h$ perpendicular to the bottom. Jumping perpendicular directs all its initial velocity against the $y'$-component of gravity, maximizing the height reached in that direction.

Q: What happens to the $x'$-component of gravity?

The $x'$-component of gravity ($-g \sin\theta$) causes the grasshopper to drift towards the lower end of the box while it's in the air.

Q: How do we know the grasshopper won't hit the side wall before escaping?

We must check the horizontal drift. As long as the drift distance is less than the width of the box, the grasshopper can choose a starting position far enough to the right to avoid hitting the left wall.

Q: Why did we approximate $45/52$ to $\sqrt{3}/2$?

In many physics problems, values are chosen to be very close to standard trigonometric ratios to allow for clean analytical answers without a calculator. $45/52 \approx 0.865$, which is extremely close to $\cos(30^\circ) \approx 0.866$.

Question

A grasshopper on the bottom of a cubical box has to jump out of the box. If each side of the box is $h = 52 	ext{ cm}$ and the grasshopper can jump with a maximum initial velocity $u = 3 	ext{ m/s}$, what should the minimum tilt angle $	heta$ the box be so that the grasshopper can jump out of the box. Acceleration due to gravity is $g = 10 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### The Grasshopper's Dilemma

Imagine you are a grasshopper trapped at the bottom of a cubical box. The box is tilted at an angle $	heta$, and you need to jump out.

Your jumping speed is limited to $u$. What is the minimum tilt angle required for you to escape?

At first glance, this looks like a complex projectile motion problem with a slanted target. But in physics, a simple shift in perspective can turn a nightmare into a walk in the park.

### Shifting the Perspective

Instead of analyzing the jump relative to the horizontal ground, let's tilt our coordinate system.

We will align our $x'$-axis along the bottom of the box and our $y'$-axis along the left wall.

In this new frame, the box is perfectly straight, but gravity is now acting at an angle!

### Resolving Gravity

Because our frame is tilted by $	heta$, the acceleration due to gravity $g$ splits into two components.

The component pulling us directly towards the bottom of the box is $g_{y'} = -g \cos	heta$.

The component pulling us towards the left wall is $g_{x'} = -g \sin	heta$.

### The Escape Strategy

To escape the box, the grasshopper must reach the top opening. In our tilted frame, this means the maximum height reached in the $y'$-direction must be at least equal to the side length of the box, $h$.

To maximize this height for a given jump speed $u$, the grasshopper must direct all its effort against the $y'$-component of gravity.

Therefore, it should jump exactly perpendicular to the bottom of the box, making its initial velocity $u_{y'} = u$.

### The Master Equation

Now, we can use the third equation of motion in the $y'$-direction:
$$v_{y'}^2 = u_{y'}^2 + 2a_{y'}y'$$

At the highest point of the jump, the vertical velocity $v_{y'}$ becomes zero. Substituting our values, we get:
$$0 = u^2 - 2(g \cos	heta)h$$

Rearranging this beautifully simple equation gives us the condition for the minimum tilt angle:
$$\cos	heta = \frac{u^2}{2gh}$$

### Final Calculation

Let's plug in the given numbers: $u = 3 	ext{ m/s}$, $g = 10 	ext{ m/s}^2$, and $h = 0.52 	ext{ m}$.

$$\cos	heta = \frac{3^2}{2 	imes 10 	imes 0.52} = \frac{9}{10.4} \approx 0.865$$

If you recall your standard trigonometric values, $\frac{\sqrt{3}}{2} \approx 0.866$.

Since the value is extremely close, we can confidently conclude that the minimum tilt angle is:
**$$	heta = 30^\circ$$**

### The Hidden Catch: Horizontal Drift

You might be wondering: what about the $x'$-component of gravity? Won't it pull the grasshopper into the left wall?

Yes, it will cause a horizontal drift $\Delta x'$ to the left. However, as long as this drift is less than the width of the box $h$, the grasshopper can simply start its jump further to the right on the bottom of the box.

For $	heta = 30^\circ$, the drift is well within the safe limit, making the escape perfectly possible!

Visualized Solution

\text{Frame of Reference}

\text{Resolving Gravity}

\text{Condition to Escape}

\text{Optimal Jump Angle}

\text{Equation of Motion}

\text{Applying the Condition}

\text{Minimum Tilt Angle}

\text{Final Calculation}

\text{What about the drift?}

The Sigma Insight: Projectile Motion

The Grasshopper's Dilemma

Shifting the Perspective

Resolving Gravity

The Escape Strategy

The Master Equation

Final Calculation

The Hidden Catch

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