A particle of mass m is projected with a velocity v making an angle of 30 degrees with the horizonta

Q: Why do we only consider the horizontal velocity at the maximum height?

At the maximum height of a projectile's trajectory, it momentarily stops moving upwards before starting to fall back down. This means its vertical velocity component is exactly zero. The only velocity it has left is the constant horizontal component, $v \cos\theta$.

Q: I tried using the cross product $\vec{r} \times \vec{v}$. Why is it so complicated?

Using the full vector cross product $(x\hat{i} + y\hat{j}) \times (v_x\hat{i} + v_y\hat{j})$ is perfectly valid and will give you the same answer! However, at the maximum height, $v_y = 0$ and $y = h$. The cross product simplifies to $x(0) - h(v_x) = -h v_x$. The geometric method ($v \cdot r_\perp$) is just a faster, more intuitive shortcut.

Q: Why is the perpendicular distance $r_\perp$ equal to $h$?

The "line of action" of the velocity at the maximum height is a perfectly horizontal line passing through that top point. The shortest, perpendicular distance from the origin $(0,0)$ to this horizontal line is simply the vertical height of the line, which is the maximum height $h$.

Q: Is the angular momentum of the projectile conserved during its flight?

No, it is not! For angular momentum to be conserved, the net external torque about the reference point must be zero. Here, gravity exerts a downward force $mg$. The position vector $\vec{r}$ and the force $m\vec{g}$ create a non-zero torque ($\vec{\tau} = \vec{r} \times m\vec{g}$), which causes the angular momentum to continuously change over time.

Q: What if the angle was $45^\circ$ instead of $30^\circ$?

The physics and the formula remain exactly the same! You would just substitute $\theta = 45^\circ$ into the final expression $L = \frac{mv^3 \sin^2\theta \cos\theta}{2g}$. Since $\sin 45^\circ = 1/\sqrt{2}$ and $\cos 45^\circ = 1/\sqrt{2}$, the result would be $\frac{mv^3}{4\sqrt{2}g}$.

Question

A particle of mass $m$ is projected with a velocity $v$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

Options:

(A) $\frac{\sqrt{3}}{2} \frac{mv^2}{g}$

(B) zero

(C) $\frac{mv^3}{\sqrt{2}g}$

(D) $\frac{\sqrt{3}}{16} \frac{mv^3}{g}$

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Flight

Imagine standing on a vast, open field and launching a particle into the air. You give it an initial velocity $v$ at an angle of $30^\circ$ to the horizontal.

As it soars through the air, it traces out a beautiful, symmetrical parabolic path. Gravity constantly pulls it downwards, slowing its vertical ascent until it reaches the very peak of its trajectory.

Our mission is to calculate the **angular momentum** of this particle about the starting point (the origin) exactly at the moment it reaches this maximum height $h$.

### The Master Equation: Demystifying Angular Momentum

Many students find angular momentum intimidating because it involves the cross product. The formal definition is $\vec{L} = \vec{r} 	imes \vec{p}$, where $\vec{r}$ is the position vector and $\vec{p} = m\vec{v}$ is the linear momentum.

However, there is a much more intuitive, geometric way to think about it. The magnitude of the angular momentum can be written as $L = m v r_\perp$.

Here, $r_\perp$ is the **perpendicular distance** from our reference point (the origin) to the "line of action" of the velocity vector. This simple geometric shortcut will save us from doing a messy vector cross product!

### The Geometry of the Highest Point

Let's apply this geometric shortcut to the highest point of the flight. What is the particle doing at the maximum height?

It has momentarily stopped moving upwards, meaning its vertical velocity is exactly zero. The only velocity it has left is its constant horizontal component, which is $v_x = v \cos 30^\circ$.

Now, imagine extending this horizontal velocity vector into an infinite line—this is the line of action. Because the particle is at height $h$, this line of action is perfectly horizontal and sits at a height $h$ above the ground.

What is the shortest, perpendicular distance from the origin $(0,0)$ to this horizontal line? It is simply the height itself! Therefore, our perpendicular distance is $r_\perp = h$.

### The Final Calculation: Bringing It All Together

Now we have everything we need. We substitute our velocity and perpendicular distance into the angular momentum formula:

$$L = m (v \cos 30^\circ) h$$

We also know the standard kinematic formula for the maximum height of a projectile:

$$h = \frac{v^2 \sin^2 30^\circ}{2g}$$

Substituting this expression for $h$ into our angular momentum equation gives us the raw setup:

$$L = m (v \cos 30^\circ) \left( \frac{v^2 \sin^2 30^\circ}{2g} ight)$$

Now, it's just a matter of careful trigonometry and algebra. We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Squaring the sine term gives us $\frac{1}{4}$.

Let's plug these numbers in:

$$L = m \left( v \frac{\sqrt{3}}{2} ight) \left( \frac{v^2 (1/4)}{2g} ight)$$

Multiplying the numerators and denominators carefully:

$$L = \frac{\sqrt{3} m v^3}{16g}$$

And there we have it! This perfectly matches option (d).

### Beyond the Answer: Is Angular Momentum Conserved?

Before we wrap up, let's think about the deeper physics. Is the angular momentum of this projectile conserved during its flight?

Absolutely not! For angular momentum to be conserved, the net external torque about the reference point must be zero.

Throughout the flight, gravity exerts a downward force $m\vec{g}$. The position vector $\vec{r}$ and this gravitational force create a non-zero torque ($\vec{	au} = \vec{r} 	imes m\vec{g}$). This torque causes the angular momentum of the particle about the origin to continuously increase as it moves forward.

Understanding *why* a quantity changes is just as important as calculating its value!

Animated Solution for Physics - Rotational Motion: A particle of mass $m$ is projected with a velocity $v$ making an angle of $3 0^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

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Visualized Solution

$Visualizing the Projectile$

$Angular Momentum Formula$

$Velocity and Perpendicular Distance$

$Substituting Maximum Height$

$Final Calculation$

$Conclusion$

$The Way Forward$

The Sigma Insight: Torque and Angular Momentum

The Setup

The Master Equation

The Geometry of the Highest Point

The Final Calculation

Beyond the Answer

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Visualized Solution

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

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

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

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

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

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

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