Question #a7296

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Question #a7296

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Answer 1 · 2017-02-07T02:13:58

$\frac{m v^{3} \sqrt{2}}{8 g}$ $(mv^3sqrt2)/(8 g)$

Explanation:

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

$\to L = \to r \times \to p$ $vecL=vecrxxvecp$
where $\to p$ $vecp$ is linear momentum and $= m \to v$ $=mvecv$
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The velocity of the particle changes continuously due to acceleration due to gravity $g$ $g$ acting on the vertical component of velocity which $= v {sin 45}^{\circ} = \frac{v}{\sqrt{2}}$ $=vsin45^@=v/sqrt2$ .
Similarly horizontal component of velocity $= v {cos 45}^{\circ} = \frac{v}{\sqrt{2}}$ $=vcos45^@=v/sqrt2$

We know that at maximum height vertical component of velocity is zero.
enter image source here
(Please read initial velocity as $v$ $v$ instead of $u$ $"u"$ in the figure, $θ = 45^{\circ}$ $theta=45^@$ )

Using the kinematic equation
$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$ and inserting given values we get
$0^{2} - {(\frac{v}{\sqrt{2}})}^{2} = 2 (- g) h$ $0^2-(v/sqrt2)^2=2(-g)h$
$\Rightarrow h = \frac{v^{2}}{2} \times \frac{1}{2 g}$ $=>h=v^2/2xx1/(2g)$
$h = \frac{v^{2}}{4 g}$ $h=v^2/(4g)$

Now $∣ ∣ ∣ \to L ∣ ∣ ∣ = h \times m ∣ ∣ \to v cos θ ∣ ∣$ $|vecL|=hxxm|vecvcostheta|$
Inserting calculated values we get

$∣ ∣ ∣ \to L ∣ ∣ ∣ = \frac{v^{2}}{4 g} \times m \times \frac{v}{\sqrt{2}}$ $|vecL|=v^2/(4g)xxmxxv/sqrt2$
$\Rightarrow ∣ ∣ ∣ \to L ∣ ∣ ∣ = \frac{m v^{3} \sqrt{2}}{8 g}$ $=>|vecL|=(mv^3sqrt2)/(8 g)$

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Question #a7296

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