Animated Solution for Physics - Rotational Motion: A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ }$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

The beauty of physics often lies in the unexpected intersection of different chapters. Here, we have a classic projectile motion setup, but we are asked to find a rotational quantity: angular momentum.

Imagine you are standing at the origin, watching a particle of mass $m$ soar through the air. It traces a perfect parabola. Our goal is to freeze time at the exact moment it reaches its maximum height, $h$, and calculate its angular momentum about your position.

To find the angular momentum $\vec{L}$ of a point particle, we use the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$:

However, dealing with vectors can sometimes be tedious. There is a much more elegant, geometric way to calculate the magnitude of angular momentum:

Here, $v$ is the speed of the particle at that instant, and $r_{\perp }$ is the perpendicular distance from our reference point (the origin) to the line of action of the velocity vector.

Let's analyze the particle at its maximum height. What is its velocity?

At the peak of a projectile's trajectory, gravity has completely halted its vertical ascent. The vertical velocity is zero! The particle is moving purely horizontally with a speed equal to the horizontal component of its initial velocity:

Now, what about $r_{\perp }$? Imagine drawing a straight, infinite line along the direction the particle is moving at the peak. This line is perfectly horizontal. If we drop a perpendicular from the origin to this horizontal line, the length of that perpendicular is exactly the maximum height of the projectile!

Therefore, our perpendicular distance is simply:

We know from kinematics that the maximum height $h$ for a projectile launched at an angle $\theta$ is given by:

Substituting $\theta =45^{\circ }$, we get:

Now comes the main point. Let's bring back our master equation and substitute these beautifully derived components:

Multiplying the numerators and denominators, we arrive at our final, elegant expression:

This perfectly matches option (b).

Before you move on, consider this thought experiment: What would be the angular momentum of the particle just before it hits the ground? The visual diagram might trick you into thinking the perpendicular distance is zero, but remember, the velocity is at an angle! The line of action does not pass through the origin. Physics is full of such beautiful subtleties. Keep questioning!