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)mgv_(0)t^(2),costhetahati`
B. `-mgv_(0)t^(2)costhetahatj`
C. `mgv_(0)tcosthetahatk`
D. `-(1)/(2)mgv_(0)t^(2)costhetahatk`

Answer

D

Explanation

The position vector of the particle at any time t during the projection is given by.
`r=v_(0)costheta t hati+(v_(0)sinthetat-(1)/(2)g t^(2))hatj`
`therefore` velocity vector, `overline(v)=(overline(dr))/(dt)`
`overline(v)=(d)/(dt)(v_(0)costheta t hati+(v_(0)sinthetat-(1)/(2)g t^(2)))hatj`
`=v_(0)costheta hati+(v_(0)sintheta-g t)hatj`
The angular momentum of the particle about the origin is
`L=m(rxxv)`
`=m[(v_(0)costhetat hati+(v_(0)sintheta t-(1)/(2)g t^(2))hatjxx(v_(0)costhetahati+(v_(0)sintheta-g t)hatj)]`
`-v_(0)^(2)sinthetacostheta t hatk +(1)/(2)v_(0)g t^(2)cos theta hatk]`
`=m[-(1)/(2)v_(0) g t^(2)costhetahatk]`
`=-(1)/(2)mgv_(0)t^(2)=costhetahatk`