A particle of mass `m` is projected with a velocity `V`. making an angle of `45^@` with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle at its maximum height h is

Options

A. zero
B. `(pi v^(3))/(4sqrt(2)g)`
C. `(piv^(3))/sqrt(2g)`
D. `m sqrt(2 gh^(3))`

Answer

B, D

Explanation

Angular momentum = (momentum) `xx` (perpendicular distance of the line of action of momentum from the axis of rotation).
Angular momentum about O,`L=(mv)/sqrt(2) xx h`……..(i)
Now, `h=(V^(2) sin^(2) theta)/(2g) = V^(2)/(4g) [therefore theta = 45^(@)]`........(ii)
From Eqs. (i) and (ii), we get: `L=m/sqrt(2)(2 sqrt(gh))h = msqrt(2gh^(3))`
Also from Eqs. (i) and (ii), we get `L =(mV)/sqrt(2) xx V^(2)/(4g) = (mV^(2))/(4sqrt(2)g)`