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A gun fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R = \(\frac{v^2_0}{g}\)

If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the fun to a height at least = ∆ [ 1 + \(\frac{∆x}{R}\) ]

(Hint: This problem can be approached in two different ways :

(i) Refer to the diagram: target T is at horizontal distance and below point of projection y = −h.

(ii) From point P in the diagram: Projection at speed at an angle θ below horizontal with height h and horizontal range ∆x.)

1 Answer

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As, R = \(\frac{v^2_0}{g}\) …(1)

Displacement along y-axis,

−ℎ = ( sin θ) + \(\frac{1}{2}\)(−g)2 …(2)

Displacement along x – axis

= \(\frac{(R+ ∆x)}{v_0cos\theta}\) …(3)

Substituting t in eq. (3),

ℎ = (−v0 sin θ) × \((\frac{R×∆x}{v_0cos\theta})+\frac{1}{2}g(\frac{R×∆x}{v_0cos\theta})^2\) 

ℎ = −(R + ∆x)tan θ + \(\frac{1}{2}g(\frac{R+∆x}{v_0cos\theta})^2\)

As θ = 45º

ℎ = −(R + ∆x)×1+\(\frac{1}{2} g\)\(\frac{R+ ∆x}{v_0^2(\frac{1}{2})}\) 

ℎ = −(R + ∆x)+ \(\frac{(R + ∆x)^2}{R}\) ∵ R = (\(\frac{v^2_0}{g}\))

= −R−∆x + (R + \(\frac{∆x^2}{R}\) + 2∆x) = ∆x +\(\frac{∆x^2}{R}\)

ℎ = ∆x(1+\(\frac{∆x^2}{R}\) ).

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