Question

Question #a7296

Answer 1

`(mv^3sqrt2)/(8 g)`

Explanation:

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

`vecL=vecrxxvecp`
where `vecp` is linear momentum and `=mvecv`

The velocity of the particle changes continuously due to acceleration due to gravity `g` acting on the vertical component of velocity which `=vsin45^@=v/sqrt2`.
Similarly horizontal component of velocity`=vcos45^@=v/sqrt2`

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

(Please read initial velocity as `v` instead of `"u"` in the figure, `theta=45^@`)

Using the kinematic equation
`v^2-u^2=2as` and inserting given values we get
`0^2-(v/sqrt2)^2=2(-g)h`
`=>h=v^2/2xx1/(2g)`
`h=v^2/(4g)`

Now `|vecL|=hxxm|vecvcostheta|`
Inserting calculated values we get

`|vecL|=v^2/(4g)xxmxxv/sqrt2`
`=>|vecL|=(mv^3sqrt2)/(8 g)`