A small paricle 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 `tlt(v_(0)sin theta)/(g)`, the angular momentum of the particle is
where `hati, hatj and hatk` are unit vectors along x, y and z-axix respectively.

Options

A. `-mgv_(0)t^(2) cos theta hatj`
B. `m g v_(0)t cos theta hatk`
C. `-1/2mgr_(0)t^(2) cos theta hatk`
D. `1/2m gv_(0)t^(2) costheta hatj`

Answer

C

Explanation

The position vector of the particles from the origin at any time t can be calculated by using `vecr = vecu t + 1/2 veca t^(2)`
`rArr vecr = v_(0) cos theta thati + (v_(0) sin theta t -1/2 g t^(2))hatj therefore` Velocity vector at any time can be calculated by using
`vecv = vecu + veca t rArr vecv = vecv_(0) costheta hati + (vecv_(0) sin theta - g t)hatj`
The angular momentum of the particles about the origin is `vecL vecr xx m vecv`
`vecL = m(r xx vecv) = m[v_(0) cos theta hati + (v_(0) sintheta t-1/2g t^(2))hatj] xx [(v_(0)cos theta hati)+ (v_(0)sin theta-g t hatj)]`
`=m[v_(0)p^(2) sin theta cos theta t hatk - v_(0)g t^(2)cos theta hatk -v_(0)^(2) sin theta cos theta hatk +1/2 v_(0)g t^(2) cos theta hatk]`
`=m[-1/2 v_(0)g t^(2) cos theta hatk] =-1/2 mgv_(0)t^(2) cos theta hatk`.