A stone is projected from the ground with a velocity 10 m/s and its trajectory is drawn to an unknow

Q: Why is the ratio of height to range the same for the actual path and the graph?

Because the graph is a uniformly scaled version of the actual trajectory. Scaling changes the size but preserves the shape and proportions, including the ratio of height to range.

Q: Why does the insect only have normal acceleration?

The problem states the insect moves with a *uniform speed*. Tangential acceleration is the rate of change of speed. Since speed is constant, tangential acceleration is zero, leaving only the normal (centripetal) component.

Q: How did we get the formula for the radius of curvature?

The formula $\rho = \frac{(1 + (y')^2)^{3/2}}{|y''|}$ is a standard result from differential calculus for any curve $y=f(x)$. It mathematically defines the radius of the circle that best approximates the curve at a given point.

Q: Why does maximum acceleration occur at the highest point?

Normal acceleration is $v^2/\rho$. Since $v$ is constant, acceleration is maximized when the radius of curvature $\rho$ is minimized. For a parabola, the sharpest turn (smallest radius) is exactly at the vertex (the highest point).

Q: Could we solve this without finding the actual range first?

Yes, but finding the actual range helps establish the scale factor explicitly, making it easier to transform the actual trajectory equation into the graph's coordinate system without getting lost in abstract variables.

Question

A stone is projected from the ground with a velocity $10 	ext{ m/s}$ and its trajectory is drawn to an unknown scale on a graph paper. The horizontal range and the maximum height on the graph are $1.0 	ext{ m}$ and $0.25 	ext{ m}$ respectively. The graph paper is glued on a horizontal tabletop. If an insect moves along the trajectory on the graph paper with a uniform speed $1.0 	ext{ cm/s}$, what should the modulus of its maximum acceleration be? Acceleration due to gravity is $10 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### The Tale of Two Trajectories

Imagine a stone soaring through the air, tracing a perfect parabola against the sky. Now, imagine someone capturing that exact motion and drawing it on a piece of graph paper.

The drawing is a perfect miniature replica, but we are not told the scale. We only know that on the paper, the horizontal range is $1.0 	ext{ m}$ and the maximum height is $0.25 	ext{ m}$.

This sets the stage for a beautiful interplay between the physical world and its mathematical representation.

### Unlocking the Angle of Projection

The genius of this problem lies in a fundamental property of projectile motion. The ratio of the maximum height $H$ to the horizontal range $R$ is entirely independent of the initial speed; it depends **only** on the angle of projection $	heta$.

Mathematically, this ratio is given by:
$$ \frac{H}{R} = \frac{	an 	heta}{4} $$

Because the drawing on the graph paper is a uniformly scaled version of the actual trajectory, this geometric ratio is perfectly preserved. We can directly use the dimensions from the graph paper to find the real-world angle!

Substituting the given values:
$$ \frac{0.25}{1.0} = \frac{	an 	heta}{4} $$

Solving this yields $	an 	heta = 1$, which means the stone was projected at exactly **$	heta = 45^\circ$**.

### Scaling the World

Now that we know the angle, we can calculate the actual horizontal range of the stone. Using the standard range formula with the given initial velocity $u = 10 	ext{ m/s}$ and $g = 10 	ext{ m/s}^2$:
$$ R = \frac{u^2 \sin 2	heta}{g} = \frac{10^2 \sin 90^\circ}{10} = 10 	ext{ m} $$

The real stone traveled $10 	ext{ m}$, but the graph paper only shows $1.0 	ext{ m}$. This immediately reveals our scale factor: $k = 0.1$. Every meter on the graph represents ten meters in reality.

Let's write the equation for the actual trajectory:
$$ y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta} $$

Plugging in our values, it simplifies elegantly to:
$$ y = x - 0.1x^2 $$

To find the equation of the curve *on the graph paper*, we apply our scale factor. Let the graph coordinates be $(X, Y)$, where $X = 0.1x$ and $Y = 0.1y$. Substituting $x = 10X$ and $y = 10Y$ into our actual equation gives:
$$ 10Y = 10X - 0.1(10X)^2 $$
$$ Y = X - X^2 $$

### The Calculus of the Crawl

Enter the insect. It is crawling along this drawn curve $Y = X - X^2$ with a constant speed of $v = 1.0 	ext{ cm/s} = 0.01 	ext{ m/s}$.

Because its speed is perfectly uniform, it has absolutely zero tangential acceleration. The only acceleration it experiences is the **centripetal (or normal) acceleration**, which forces it to turn along the curve.

This acceleration is given by:
$$ a_n = \frac{v^2}{ho} $$
where $ho$ is the radius of curvature.

To maximize the acceleration, we must minimize the radius of curvature. From differential calculus, the radius of curvature for any function $Y(X)$ is:
$$ ho = \frac{(1 + (Y')^2)^{3/2}}{|Y''|} $$

Let's find the derivatives of our graph equation $Y = X - X^2$:
$$ Y' = 1 - 2X $$
$$ Y'' = -2 $$

### The Final Calculation

Substituting these derivatives into our radius of curvature formula:
$$ ho = \frac{(1 + (1 - 2X)^2)^{3/2}}{|-2|} $$

To make $ho$ as small as possible, the squared term $(1 - 2X)^2$ must be zero. This happens at $X = 0.5$, which is the exact peak of the parabola!

At this peak, the minimum radius of curvature is:
$$ ho_{	ext{min}} = \frac{(1 + 0)^{3/2}}{2} = 0.5 	ext{ m} $$

Finally, we calculate the maximum acceleration of the insect:
$$ a_{	ext{max}} = \frac{v^2}{ho_{	ext{min}}} = \frac{(0.01)^2}{0.5} = \frac{10^{-4}}{0.5} = 2 	imes 10^{-4} 	ext{ m/s}^2 $$

And there we have it! A stunning synthesis of kinematics, scaling geometry, and differential calculus leading to a beautifully precise answer.

Visualized Solution

$Understanding the Setup$

$Angle of Projection$

$Calculating the Angle$

$Actual Range and Scale Factor$

$Equation of Actual Trajectory$

$Equation on Graph Paper$

$Acceleration of the Insect$

$Radius of Curvature$

$Minimum Radius of Curvature$

$Maximum Acceleration$

The Sigma Insight: Projectile Motion

The Tale of Two Trajectories

Unlocking the Angle of Projection

Scaling the World

The Calculus of the Crawl

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle of Projection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Actual Range and Scale Factor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Actual Trajectory

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation on Graph Paper

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration of the Insect

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of Curvature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Minimum Radius of Curvature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Acceleration