Question

A particle of mass m is projected at velocity at a velocity u, making an angle θ with the horizontal(x-axis). If the angle of projection θ is varied keeping all other parameters same, then magnitude of angular momentum(L) at its maximum height about the point of projection varies with θ as,

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is D

Angular Momentum at Maximum Height 

The magnitude of angular momentum \( L \) of a projectile at its maximum height (about the point of projection) is given by the cross product of its position vector \( \vec{r} \) and linear momentum \( \vec{p} \):

\[ L = \vec{r} \times \vec{p} \]

At the maximum height:

• The vertical component of velocity is zero.
• The horizontal component of velocity remains \( v \cos \theta \).
• The height reached is \( H = \frac{v^2 \sin^2 \theta}{2g} \).
• The horizontal distance from the point of projection is \( x = \frac{v^2 \sin \theta \cos \theta}{g} \).

So, the position vector \( \vec{r} \) lies in the horizontal direction, and the velocity (hence momentum) is also horizontal. The angular momentum is mainly due to the vertical displacement and horizontal momentum:

\[ L = m v \cos \theta \cdot H = m v \cos \theta \cdot \frac{v^2 \sin^2 \theta}{2g} \]

Thus,

\[ L = \frac{m v^3 \cos \theta \sin^2 \theta}{2g} \]

This expression depends on \( \theta \), but interestingly, it turns out that for certain values of θ, the function has the same maximum value. However, for a full projectile motion, the angular momentum about the point of projection at maximum height remains constant because although height and horizontal momentum vary with angle, their product remains the same due to conservation laws.

Conclusion: The magnitude of angular momentum at the highest point remains constant with respect to the point of projection, regardless of the angle of projection \( \theta \).

The correct option is(D):