A small particle of mass ` m` is projected at an angle `theta` with the ` x`- axis with an initial velocity ` v_(0)` in the ` x-y `plane as shown in the figure . At a time ` t lt ( v_(0) sin theta)/(g) `, the angular momentum of the particle is
where `hat (i) , hat (j) and hat(k)` are unit vectors along `x , y and z` - axis respectively.

Options

A. `(1)/(2) mg v_(0) t^(2)cos theta hati`
B. `-mg v_(0) t^(2)cos theta hati`
C. `mg v_(0) t cos theta hatk`
D. `-(1)/(2)mgv_(0) t^(2) cos theta hatk`

Answer

D

Explanation

Let the particle reach at point Pin time t. Horizontal distance travelled by particle is `(v_(0)costheta)t` which is perpendicular to the line of action of force acting on the particle as shown in figure,
Torque acting about .O. is:
Torque = force `xx` perpendicular distance of line of action of force
`tau=mg(v_(0)costheta)t`
Also, `vectau=(dvecL)/(dt)`
`:.vecL=(int_(0)^(t)mg(v_(0)costheta)tdt)hatk`
`vecL=-(mgv_(0)costhetat^(2))/(2)hatk`
For motion of the particle in X-Y plane, `vecL` is along `-hatk`
`:.vecL=-(mgv_(0)costhetat^(2))/(2)hatk`