4.30. A gun can fire shells with maximum speed v0 and the maximum horizontal range that can be achieved is R=v20g. If a target farther away by distance ∆x has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least h=Δx[1+ΔxR]
This problem can be approached in two different ways
(i) Refer to the diagram, target T is at a horizontal distance x=R+Δx and between the point of projection y = - h.
(ii) From point P in the diagram projection at speed v0 at an angle θ below horizontal with height h and horizontal range (∆x)
Hint: Time is taken to travel horizontal and vertical distances will be the same.
Step 1: Find the maximum range.
R = v20g, for θ = 45∘ ...(i)
Step 2: Apply equation in the horizontal direction
The horizontal component of initial velocity =v0cosθ
(R+Δx)=v0cosθ×t⇒ t=(R+Δx)v0cosθ ...(ii)
Step 3: Put the value of t in from equation (ii) for vertical motion and calculate h
Let the gun be raised through a height h from the ground so that it can hit the target. Let vertically downward direction is taken as positive
The vertical component of initial velocity =−v0sinθ
Taking motion in the vertical direction, h=(−v0sinθ)t+12gt2 ...(iii)
Substituting the value of t in Eq. (iii), we get
h=(−v0sinθ)×(R+Δxv0cosθ)+12g(R+Δxv0cosθ)2h=−(R+Δx)tanθ+12g(R+Δx)2v20cos2θ
As the angle of projection is θ = 45°, therefore
h=−(R+Δx)+tan45∘+12g(R+Δx)2v20cos245∘h=−(R+Δx)×1+12g(R+Δx)2v20(1/2)
(∵ tan45∘=1 and cos45∘=1√2)
h=−(R+Δx)+(R+Δx)2R [using EQ. (i) R = v20/g]=−(R+Δx)+1R(R2+Δx2+2RΔx)=−R−Δx+(R+Δx2R+2Δx)=Δx+Δx2Rh=Δx(1+ΔxR)
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