# Question #a7296

Feb 7, 2017

$\frac{m {v}^{3} \sqrt{2}}{8 g}$

#### Explanation:

We know that angular momentum $\vec{L}$ of a particle of mass $m$ moving with velocity $\vec{v}$ with respect to a selected origin is given by the equation

$\vec{L} = \vec{r} \times \vec{p}$
where $\vec{p}$ is linear momentum and $= m \vec{v}$

The velocity of the particle changes continuously due to acceleration due to gravity $g$ acting on the vertical component of velocity which $= v \sin {45}^{\circ} = \frac{v}{\sqrt{2}}$.
Similarly horizontal component of velocity$= v \cos {45}^{\circ} = \frac{v}{\sqrt{2}}$

We know that at maximum height vertical component of velocity is zero.

(Please read initial velocity as $v$ instead of $\text{u}$ in the figure, $\theta = {45}^{\circ}$)

Using the kinematic equation
${v}^{2} - {u}^{2} = 2 a s$ and inserting given values we get
${0}^{2} - {\left(\frac{v}{\sqrt{2}}\right)}^{2} = 2 \left(- g\right) h$
$\implies h = {v}^{2} / 2 \times \frac{1}{2 g}$
$h = {v}^{2} / \left(4 g\right)$

Now $| \vec{L} | = h \times m | \vec{v} \cos \theta |$
Inserting calculated values we get

$| \vec{L} | = {v}^{2} / \left(4 g\right) \times m \times \frac{v}{\sqrt{2}}$
$\implies | \vec{L} | = \frac{m {v}^{3} \sqrt{2}}{8 g}$