Question

A particle of mass m is projected with velocity v making an angle of 45º with the horizontal. The magnitude of the angular momentum of projectile about the axis of projection when the particle is at maximum height is
\[\begin{align}
  & \text{A}\text{. Zero} \\
 & \text{B}\text{. }\dfrac{m{{v}^{3}}}{4\sqrt{2}g} \\
 & \text{C}\text{. }\dfrac{m{{v}^{3}}}{\sqrt{2}g} \\
 & \text{D}\text{.}\dfrac{m{{v}^{3}}}{4g} \\
\end{align}\]

Answer 1

Hint: Using angular momentum formula we can solve this problem.

Formula used: Angular momentum \[L=\dfrac{mv}{\sqrt{2}}\dfrac{{{v}^{2}}}{4g}=\dfrac{m{{v}^{3}}}{4\sqrt{2}g}\]where, m = mass, v = velocity and h = height.
And, maximum height \[h=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] where,= the angle with the horizontal.

Complete step by step solution:
Let us consider v as the velocity of the projection and angle of projection given is 45º.
\[L=mvr\sin \theta \]
At maximum point, velocity \[v=v\cos \theta =\dfrac{1}{\sqrt{2}}v\] direction towards horizontal and no vertical velocity is present).
The maximum height reached will be \[h=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}=\dfrac{{{v}^{2}}{{\sin }^{2}}\left( {{45}^{\circ }} \right)}{2g}=\dfrac{{{v}^{2}}}{4g}............\left( ii \right)\]
Now, let us substitute equation (ii) in (i) we get,
\[L=\dfrac{mv}{\sqrt{2}}\dfrac{{{v}^{2}}}{4g}=\dfrac{m{{v}^{3}}}{4\sqrt{2}g}\]

Therefore, the required answer is \[\dfrac{m{{v}^{3}}}{4\sqrt{2}g}\]i.e., option B.

Note: Angular momentum is an important quantity in physics as it is a conserved quantity. The angular momentum is the rotational equivalent of linear momentum.