A gun is fired from a moving platform and ranges of the shot are observed to be `R_1 and R_2` when the platform is moving forwards and backwards, respectively, with velocity `v_P`. Find the elevation of the gun `prop` in terms of the given quantities.

Explanation

For projectile:
Horizontal component `=u cos theta+v_(0)`
vertical component `= u sin theta`
`T=(2u sin theta)/g........(i)`
`R=(u cos theta +v_(0))T........(ii)`

Horizontal component `=u cos theta -v_(0)`
`R_(0)=(u cos theta-v_(0))T......(iii)`
Dividing (ii) by (iii), we get
`R/(R_(0)) =(u cos theta+v_(0))/(u cos theta-v_(0))`
`u cos theta=((R+R_(0))v_(0))/(R-R_(0)).........(iv)`
`R(u cos theta+v_(0))T`
`=[((R+R_(0))v_(0))/(R-R_(0))+v_(0)](2 u sin theta)/g`
`u sin theta=(g(R-R_(0)))/(4v_(0))........(v)`
Dividing (v) by (iv), we get
`tan theta=(g(R-R_(0))^(2))/(4v_(0)^(2)(R+R_(0)))`
`theta=tan^(-1)[(g(R-R_(0))^(2))/(4v_(0)^(2)(R+R_(0)))]`