A shell is projected from a gun with a muzzle velocity `v`.The gun is fitted with a trolley car at an angle `theta` as shown in the fig. If the trolley car is made to move with constant velocity `v` towards right, find the horizontal range of the shell relative to ground.

Options

A. `R=(2u sin theta(u cos theta+v))/g`
B. `R=(2u sin theta(u cos theta-v))/g`
C. `R=(u sin theta(u cos theta+v))/g`
D. `R=(u sin theta(u cos theta-v))/g`

Answer

A

Explanation

The velocity of projection of the shell is
`vecv_(s)=vecv_(sc)=vecv_(c)`.Substituting
`vec v_(sc)= u cos theta hati+u sin thetahatj` and `vecv_(c)-v hatj`
For horizontal range `R` of the shell its dispalcement can be given as `vecs=Rhati` Substituting
`veca=-g, vecs=Rhati`,and `vecu=vecv_(s)`
`=(u cos theta +v)+u sin theta hatj "in" vecs=vecut +1/2vecat^(2)`
`Rhati=(u cos theta+v)thati+(ut sin theta-1/2"gt"^(2))hatj`
Compating the coefficients of `hati` and `hatj` , we obtain
`R=(u cos theta+v)t`...(i) and
`ut sin theta-1/2"gt"^(2)=0`..(ii) From eq. (ii), ltbgt we find `t=(2u sin theta)/g` Finally, substituting
`t=(2u sin theta)/g` eq.(i), we have
`R=(2u sin theta(u cos theta+v))/g`