A particle of mass m is projected with a velocity ' u ' 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 :

Rotational Motion PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q64 with Solution

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Correct answer: D

Given data

A particle is projected with speed $u$ at an angle $30^\circ$ above the horizontal.

We need the magnitude of angular momentum about the point of projection when the projectile is at its maximum height.

Angular momentum about origin:

\vec L = \vec r \times m\vec v

So magnitude is

L = mrv\sin\phi

where $\phi$ is the angle between $\vec r$ and $\vec v$ .

Velocity at maximum height

At the highest point, the vertical component of velocity becomes zero. So only horizontal component remains:

v = u\cos 30^\circ = \frac{\sqrt{3}}{2}u

Direction of velocity is horizontal.

Coordinates of maximum height

Maximum height of projectile is

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

Since $\sin 30^\circ = \frac12$ ,

h = \frac{u^2\left(\frac14\right)}{2g} = \frac{u^2}{8g}

At maximum height, position vector from point of projection has vertical coordinate $h$ .

Perpendicular distance method

Since velocity at top is horizontal, the angular momentum magnitude about the point of projection is

L = m v \times (\text{perpendicular distance from origin to line of motion})

The line of motion at the top is horizontal through height $h$ , so perpendicular distance from origin is simply $h$ .

Thus,

L = m\left(u\cos30^\circ\right)h

Substitute values:

L = m\left(\frac{\sqrt{3}}{2}u\right)\left(\frac{u^2}{8g}\right)

L = \frac{\sqrt{3}}{16}\frac{mu^3}{g}

Match with options

This matches:

\boxed{\frac{\sqrt{3}}{16}\frac{mu^3}{g}}

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

More from Rotational Motion

View written solutionFree

Correct answer: D

Given data

A particle is projected with speed $u$ at an angle $30^\circ$ above the horizontal.

We need the magnitude of angular momentum about the point of projection when the projectile is at its maximum height.

Angular momentum about origin:

\vec L = \vec r \times m\vec v

So magnitude is

L = mrv\sin\phi

where $\phi$ is the angle between $\vec r$ and $\vec v$ .

Velocity at maximum height

At the highest point, the vertical component of velocity becomes zero. So only horizontal component remains:

v = u\cos 30^\circ = \frac{\sqrt{3}}{2}u

Direction of velocity is horizontal.

Coordinates of maximum height

Maximum height of projectile is

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

Since $\sin 30^\circ = \frac12$ ,

h = \frac{u^2\left(\frac14\right)}{2g} = \frac{u^2}{8g}

At maximum height, position vector from point of projection has vertical coordinate $h$ .

Perpendicular distance method

Since velocity at top is horizontal, the angular momentum magnitude about the point of projection is

L = m v \times (\text{perpendicular distance from origin to line of motion})

The line of motion at the top is horizontal through height $h$ , so perpendicular distance from origin is simply $h$ .

Thus,

L = m\left(u\cos30^\circ\right)h

Substitute values:

L = m\left(\frac{\sqrt{3}}{2}u\right)\left(\frac{u^2}{8g}\right)

L = \frac{\sqrt{3}}{16}\frac{mu^3}{g}

Match with options

This matches:

\boxed{\frac{\sqrt{3}}{16}\frac{mu^3}{g}}

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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