A gun is mounted on a trolley which can move uniformly with speed `v m//s` along the `x`-axis. Two shots are find from the origin with the gun making an angle `30^(@)` with the horizontal such that in the first case the trolley is moving along the positive `x`-axis and in the second case moving along the negative `x`- axis. The respective range of the projectile is `250 m` and `200 m`, along the `x-`axis. Find the velocity of the trolley in `m//s`. (Assume height of the trolley to be negligible)(given `g=10m//s^(2)`).

Explanation

Let `u` be the velocity of the projectile w.r.t. the trolley.
Case I : The trolley moves forward
horizontal velocity `U_(x)=ucos30^(@)+v`
vertical velocity `U_(y)=usin30^(@)`
Time of light `T=(2usin30^(@))/(g)`
Range `R_(1)=U_(x)T=(2usin30^(@))/(g)[ucos30^(@)+v]`....`(1)`
Case II : The trolley moves backward
horizontal velocity `U_(x)=ucos30^(@)-v`
vertical velocity `U_(y)=u sin30^(@)`
Range `R_(2)=U_(x)T=(2usin30^(@))/(g)[ucos30^(@)-v]`.....`(2)`
Adding and subtracting `(1)` and `(2)`,
`R_(1)+R_(2)=(4u^(2))/(g)(sin30^(@)cos30^(@))`
`R_(1)-R_(2)=(4uvsin30^(@))/(g)`
Eliminating `u` from the two equations
`v^(2)=(g)/(4tan30^(@))((R_(1)-R_(2))^(2))/((R_(1)+R_(2)))`
here, `R_(1)=250m` , `R_(2)=200 m` , `g=10m//s^(2)`
`v^(2)=25m^(2)//s^(2)` or `v=5m//s`.