A gun can fire shells with maximum speedv_0 and the maximum horizontal range that can be achieved isR = v_0^2 //g If a target farther away by distanceDelta x (beyound R) has to be hit with the asme gun 9Fig. 2 (EP).24). Show that it could be accieved by raising the gun to a height at least h= Delta x [ 1 + (Delta x0/R]. .

A gun can fire shells with maximum speedv_0 and the maximum horizontal

1.	A gun can fire shells with maximum speedv_0 and the maximum horizontal range that can be achieved isR = v_0^2 //g If a target farther away by distanceDelta x (beyound R) has to be hit with the asme gun 9Fig. 2 (EP).24). Show that it could be accieved by raising the gun to a height at least h= Delta x [ 1 + (Delta x0/R]. .
Answer» Solution :As per wuestion, the maximu , bborizontal range on ground is ` R_(max) = (v_0&2 /g` It isif ` 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 + 1/2t+ 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_0cos 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 + Deltax) 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)`.