A body is projected at t=0 with a velocity 10 ms^-1 at an angle of 60 degrees with the horizontal. T

Q: Why is the normal acceleration $g \cos\alpha$ and not $g \sin\alpha$?

Draw a diagram! The velocity vector makes an angle $\alpha$ with the horizontal. The normal direction is perpendicular to the velocity, so it makes an angle $\alpha$ with the vertical. Since gravity acts purely vertically, its component along the normal is $g \cos\alpha$.

Q: How are we supposed to know that $\tan\alpha = 2 - \sqrt{3}$ means $\alpha = 15^\circ$?

This is a standard trigonometric value that frequently appears in JEE problems. You can derive it using the half-angle formula: $\tan(30^\circ/2) = \frac{1 - \cos 30^\circ}{\sin 30^\circ} = \frac{1 - \sqrt{3}/2}{1/2} = 2 - \sqrt{3}$. Memorizing values for $15^\circ$ and $75^\circ$ will save you a lot of time!

Q: Why did we take the absolute value of $v_y$ when calculating $\tan\alpha$?

The negative sign in $v_y$ simply indicates direction (downwards). By taking the absolute value, we are finding the geometric angle $\alpha$ below the horizontal, which makes it easier to visualize the right triangle for the normal acceleration component.

Q: Can I use the mathematical formula for radius of curvature $R = \frac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|}$?

Yes, absolutely! You can find the equation of the trajectory $y(x)$, calculate the first and second derivatives at the specific $x$ coordinate corresponding to $t=1$, and plug them into the formula. However, the kinematic approach ($v^2/a_n$) is usually much faster and less prone to algebraic errors in physics problems.

Q: What does it mean physically when the radius of curvature changes?

A smaller radius of curvature means the path is turning very sharply, while a larger radius means the path is relatively flat or straight. In projectile motion, the path is sharpest (minimum $R$) at the highest point where the speed is lowest and the entire gravity vector acts as normal acceleration.

Question

A body is projected at $t=0$ with a velocity $10 	ext{ ms}^{-1}$ at an angle of $60^\circ$ with the horizontal. The radius of curvature of its trajectory at $t=1	ext{ s}$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g = 10 	ext{ ms}^{-2}$, the value of $R$ is

Options:

(A) $10.3 	ext{ m}$

(B) $2.8 	ext{ m}$

(C) $5.1 	ext{ m}$

(D) $2.5 	ext{ m}$

Sigma Percentile · Accepted Answer

Have you ever watched a baseball fly through the air and wondered about the exact shape of its curve? While we know it's a parabola, the "sharpness" of that curve changes at every single point. This sharpness is mathematically described by the **radius of curvature** ($R$).

In this thrilling problem, we are asked to find the radius of curvature of a projectile exactly 1 second after it is launched. Let's dive into the beautiful kinematics and geometry that make this possible!

### The Master Formula

The radius of curvature $R$ at any point on a trajectory is intimately tied to the object's speed and the acceleration pulling it sideways (perpendicular to its motion). The relationship is given by:
$$R = \frac{v^2}{a_n}$$
where $v$ is the instantaneous speed and $a_n$ is the **normal component of acceleration**.

For a projectile in free fall, the only acceleration is gravity ($g$) acting straight down. If the velocity vector makes an angle $\alpha$ with the horizontal, the component of gravity perpendicular to the velocity is:
$$a_n = g \cos\alpha$$

So, our mission is clear: find the velocity $v$ and the angle $\alpha$ at $t = 1 	ext{ s}$.

### Breaking Down the Velocity

Let's start by splitting the initial velocity into its horizontal and vertical components. The projectile is launched at $10 	ext{ m/s}$ at an angle of $60^\circ$.

**Horizontal Velocity ($v_x$):**
Gravity doesn't pull sideways, so the horizontal velocity remains perfectly constant throughout the flight.
$$v_x = u \cos 60^\circ = 10 	imes \frac{1}{2} = 5 	ext{ m/s}$$

**Vertical Velocity ($v_y$):**
The vertical velocity starts upward but is constantly reduced by gravity. After 1 second, we have:
$$v_y = u \sin 60^\circ - gt = 10 	imes \frac{\sqrt{3}}{2} - 10(1) = 5\sqrt{3} - 10 	ext{ m/s}$$
Since $5\sqrt{3} \approx 8.66$, the vertical velocity is negative ($-1.34 	ext{ m/s}$). This tells us the projectile has already passed its peak and is on its way down!

### The Angle of Descent

Now, we need the angle $\alpha$ that the velocity vector makes with the horizontal. We can find this using the tangent function:
$$	an\alpha = \frac{|v_y|}{v_x} = \frac{10 - 5\sqrt{3}}{5} = 2 - \sqrt{3}$$

Here is where a bit of mathematical elegance comes in. The value $2 - \sqrt{3}$ is a standard trigonometric identity for $15^\circ$. Therefore:
$$\alpha = 15^\circ$$

### The Final Calculation

We have everything we need! First, let's find the square of the total speed ($v^2$) by adding the squares of the components:
$$v^2 = v_x^2 + v_y^2 = 5^2 + (10 - 5\sqrt{3})^2$$
$$v^2 = 25 + 100 + 75 - 100\sqrt{3} = 200 - 100\sqrt{3} = 100(2 - \sqrt{3})$$

Now, we plug $v^2$ and $a_n = g \cos 15^\circ$ into our master formula:
$$R = \frac{100(2 - \sqrt{3})}{10 \cos 15^\circ}$$

Using $\sqrt{3} \approx 1.732$ and $\cos 15^\circ \approx 0.966$:
$$R = \frac{100(0.268)}{10 	imes 0.966} = \frac{26.8}{9.66} \approx 2.77 	ext{ m}$$

Rounding to one decimal place, we get our final answer: **$2.8 	ext{ m}$**.

The geometry of the curve at that exact moment corresponds to a circle with a radius of 2.8 meters. Isn't it amazing how simple kinematic equations can reveal such deep geometric truths?

Select Answer:

Visualized Solution

$Visualizing the Trajectory$

$Radius of Curvature Formula$

$Velocity Components$

$Calculating v_{x}$

$Calculating v_{y}$

$Angle of Velocity α$

$Square of Velocity v^{2}$

$Final Calculation$

The Sigma Insight: Projectile Motion

The Master Formula

Breaking Down the Velocity

The Angle of Descent

The Final Calculation

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Select Answer:

Visualized Solution

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Square of Velocity v2

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