An object of mass ' $m$ ' is projected from origin in a vertical $x y$ plane at an angle $45^{\circ}

Question

An object of mass ' $m$ ' is projected from origin in a vertical $x y$ plane at an angle $45^{\circ}$ with the $x$-axis with an initial velocity $v_{0}$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ $g$ is acceleration due to gravity]\

Professor PC Thomas Classes · Answer

(d) : Given, mass of object $m$ Initial velocity $=v_{0}$\ Angle of projection $=45^{\circ}$\ ∵ Horizontal component of velocity, $$ 
u_{x}=
u_{0} \cos 	heta $$ Vertical component of velocity, $$ 
u_{y}=
u_{0} \sin 	heta $$ Maximum height, $H=\frac{v_{0}^{2} \sin ^{2} 	heta}{2 g}$ $$ =\frac{v_{0}^{2} \sin ^{2} 45^{\circ}}{2 g} \Rightarrow H_{\max }=\frac{v_{0}^{2}}{4 g} $$ Horizontal momentum, $$ p_{x}=m v_{0} \cos 45^{\circ}=\frac{m v}{\sqrt{2}} $$ Angular momentum, $L=m v r_{\perp}$ $$ =\frac{m v_{0}}{\sqrt{2}} 	imes \frac{v_{0}^{2}}{4 g} ; L=\frac{m v_{0}^{3}}{4 \sqrt{2} g} $$ The direction of angular momentum is along $m(\vec{v} 	imes \vec{r})$ that is perpendicular to the $x-y$ plane and in the negative $z$-direction.\