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

As per wuestion, the maximu , bborizontal range on ground is ` R_(max) = (v_0&2 /g`
It is if ` theta = 54^@` . The shell fired at (0) from height (h) can hit the same target at B. Fig. 2 ( EF). 25 `
Taking vertical downward motion from (O) to (B), we have `
` u_y v_0 sin theta , a_y = g, y_0 =0 , y=h, t=?`
as, ` y= y_0 + u_y + 1/2 a_y t^2`
:. ` x=x _0 + u_x + 1/2 t+ a_x t^2`
(R+ Delta x) =0 + v) cos thea t` or ` t= ((R + delta x))./(v_0 cos theta)`
Putting value fo (t) in (i), we have
` h=- v_0 sin theta xx (R + Delta x)/(v_0 cos theta) + 1/2 g ((R + Delta x^2)/(v_0^2 cos theat ) =- (R + Delat x) tan theta + 1/2 g ((R + Detal x)^2)/(v_0^2 cos theta^2 theta)`
As ` theta = 45^@` , so ` h=- (R + Delta x) tan 45^@ + 1/2 (g(R+ Delta x)^2)/( 2 v_0^2 cos^@ 45^@`
or ` h=- (R + Delta x) xx 1 + 1/2 (g(R+ Delta x)^2)/v_0^2` xx 2 =- (R + Delta x) + 1/ R xx (R + Delta x)^2` [:. v_0^2 //g = R]`
` =- (R + Delta x) + R + 2 Delta x + Delta x^2 //R = Delta x + Delta x^2 //R = Delta x (1 + Delta x//R)`.