A gun can fire shells with maximum speed `v_(0)` and the maximum horizontal range that can be achieved is `R=(v_(0)^(2))/(g)`. If a target farther away by distance `Deltax` (beyond R) has to be hit with the same gun, show that it could that it could be achieved by raising the gun to a height at least `h=Delta x[1+(Deltax)/(R)]`

Explanation

This problem can be approached in two different ways
(i) Refer to the diagram , target T is at horizontal distance `x=R+Deltax` and between point of projection y= -h.
(ii) From point P in the diagram projection at speed `v_(0)` at an angle `theta` below horizontal with height h and horizontal range `DeltaxA)`
Applying method (i)
Maximum horizontal range
`R=(v_(0)^(2))/(g) , "for" theta=45^(@)" " ...(i)`
Let the gun be raised through a height h from the ground so that it can hit the traget. Let vertically downward direction is taken as positive
Horizontal component of initial velocity `=v_(0) cos theta`
Vertical component of initial velocity `=-v_(0) sin theta`
Taking motion in vertical direction, `h=(-v_(0)sin theta)t=(1)/(2) "gt"^(2) " " ...(ii)`
Taking motion in horizontal direction
`(R+Deltax)=v_(0)costhetaxxt`
`rArr" " t=((R+Deltax))/(v_(0)costheta) " " ...(iii)`
Substituting value of t in Eq. (ii) , we get
`h=(-v_(0)sin theta)xx((R+Deltax)/(v_(0)cos theta)) + (1)/(2) g ((R+Deltax)/(v_(0)cos theta))^(2)`
`h=-(R+Deltax)tan theta + (1)/(2) g ((R+Deltax)^(2))/(v_(0)^(2)cos^(2)theta)`
As angle of projection is `theta = 45^(@)` , therefore
`h=-(R+Deltax)+tan 45^(@) + (1)/(2) g. ((R+Deltax)^(2))/(v_(0)^(2)cos^(2)45^(@))`
`h=-(R+Deltax)xx1+(1)/(2)g.((R+Deltax)^(2))/(v_(0)^(2)(1//2))`
`" " (.:.tan 45^(@)=1"and" cos 45^(@)=(1)/(sqrt(2)))`
`h=-(R+Deltax)+((R+Deltax^(2)))/(R) " "` [Using Eq. (i) , `R=v_(0)^(2)//g`]
`=-(R+Deltax)+(1)/(R) (R^(2)+Deltax^(2)+2R Deltax)`
`=-R-Delta x + (R+(Deltax^(2))/(R)+2Deltax)`
`=Deltax + (Deltax^(2))/(R)`
`h=Deltax(1+(Deltax)/(R))`