Question

A particle of mass \(m\) is projected with speed \(v\) at \(45^{\circ }\) with the horizontal. The magnitude of angular momentum of the particle about the point of projection when the particle is at its maximum height \(h\) is \(\left(\cos 45^{\circ }=\frac{1}{\sqrt{2}}\right)\)

• A. zero
• B. \(\frac{mvh^{2}}{\sqrt{2}}\)
• C. \(\frac{mv^{2}h}{\sqrt{2}}\)
• D. \(\frac{mvh}{\sqrt{2}}\)

Hint:

At the highest point only the horizontal velocity remains and the position vector from the point of projection is vertical, so the angular momentum is mass times horizontal speed times height.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Angular momentum about the point of projection is \(L=mvr\sin\phi\), which for perpendicular velocity and position components is \(m\,v_{x}\,y\).

Step 2: Values at the Top
At the highest point the speed is \(v_{x}=v\cos 45^{\circ}=\dfrac{v}{\sqrt{2}}\) and the vertical distance from the point of projection is \(h\).

Step 3: Calculation
\(L=m\times\dfrac{v}{\sqrt{2}}\times h=\dfrac{mvh}{\sqrt{2}}\)

Step 4: Compare the options
This matches option (D).

Final Answer:
The angular momentum is \(\dfrac{mvh}{\sqrt{2}}\). \[ \boxed{\text{(D) }\dfrac{mvh}{\sqrt{2}}} \]