Question #a7296

1 Answer
Feb 7, 2017

(mv^3sqrt2)/(8 g)mv328g

Explanation:

We know that angular momentum vecLL of a particle of mass mm moving with velocity vecvv with respect to a selected origin is given by the equation

vecL=vecrxxvecpL=r×p
where vecpp is linear momentum and =mvecv=mv
![wikimedia.org](useruploads.socratic.org)

The velocity of the particle changes continuously due to acceleration due to gravity gg acting on the vertical component of velocity which =vsin45^@=v/sqrt2=vsin45=v2.
Similarly horizontal component of velocity=vcos45^@=v/sqrt2=vcos45=v2

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

Using the kinematic equation
v^2-u^2=2asv2u2=2as and inserting given values we get
0^2-(v/sqrt2)^2=2(-g)h02(v2)2=2(g)h
=>h=v^2/2xx1/(2g)h=v22×12g
h=v^2/(4g)h=v24g

Now |vecL|=hxxm|vecvcostheta|L=h×mvcosθ
Inserting calculated values we get

|vecL|=v^2/(4g)xxmxxv/sqrt2L=v24g×m×v2
=>|vecL|=(mv^3sqrt2)/(8 g)L=mv328g