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Chapter 2: Problem 69

If the range of a gun which fires a shell with muzzle speed \(V\) is \(R\), then the angle of elevation of the gun is (A) \(\cos ^{-1}\left(\frac{V^{2}}{R g}\right)\) (B) \(\cos ^{-1}\left(\frac{g R}{V^{2}}\right)\) (C) \(\frac{1}{2}\left(\frac{V^{2}}{R g}\right)\) (D) \(\frac{1}{2} \sin ^{-1}\left(\frac{g R}{V^{2}}\right)\)

Short Answer

Expert verified
The angle of elevation of the gun is \(\theta = \frac{1}{2}\sin^{-1} \left(\frac{gR}{V^2}\right)\).

Step by step solution

01

Find the range equation in terms of the angle

Use the horizontal and vertical components of the projectile motion to find the range equation: For a projectile, the range \(R\) is given by: \[R = \frac{V^2 \sin 2\theta}{g}\] Where \(V\) is the initial speed, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of elevation. We need to find the \(\theta\) expression from the given options.
02

Rearrange the range equation for the angle

We will rearrange the range equation to isolate the angle term: \[\sin 2\theta = \frac{gR}{V^2}\] Now, take the inverse sine on both sides to get the angle: \[2\theta = \sin^{-1} \left(\frac{gR}{V^2}\right)\]
03

Divide by 2 to find the angle of elevation

Finally, divide both sides by 2 to find the angle of elevation: \[\theta = \frac{1}{2}\sin^{-1} \left(\frac{gR}{V^2}\right)\] The answer is (D).

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Most popular questions from this chapter

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