91Ó°ÊÓ

To prove whether or not \(\alpha + \beta = \frac{\pi}{2}\), we will start by finding an expression for the ranges of both projectiles: Projectile 1: \(R_1 = \frac{u^2 \sin 2\alpha}{g}\) Projectile 2: \(R_2 = \frac{u^2 \sin 2\beta}{g}\) Since both projectiles have the same range, we can write: \(\frac{u^2 \sin 2\alpha}{g} = \frac{u^2 \sin 2\beta}{g}\), Simplifying, we have: \(\sin 2\alpha = \sin 2\beta\) Applying the sum-to-product formula: \(\sin (2\alpha - 2\beta) = 0\) Since \(\sin x = 0\) when \(x = n\pi\), where n is any integer, we get: \(2\alpha - 2\beta = n\pi\) Dividing both sides by 2 yields: \(\alpha - \beta = \frac{n\pi}{2}\) This does not prove that \(\alpha + \beta = \frac{\pi}{2}\), so choice A is false.

We'll start by finding expressions for the maximum heights: Projectile 1: \(h_1 = \frac{u^2 \sin^2 \alpha}{2g}\) Projectile 2: \(h_2 = \frac{u^2 \sin^2 \beta}{2g}\) The given relationship is: \(R=4\sqrt{h_1 h_2}\) Substituting \(R_1\) and our expressions for \(h_1\) and \(h_2\), we have: \(\frac{u^2\sin 2\alpha}{g}=4\sqrt{\frac{u^2 \sin^2 \alpha}{2g} \cdot \frac{u^2 \sin^2 \beta}{2g}}\) Canceling out some common terms, we get: \(\sin 2\alpha = 4\sin\alpha\sin\beta\) Applying the double-angle formula for sine, we get: \(2\sin\alpha\cos\alpha = 4\sin\alpha\sin\beta\) Dividing both sides by \(2\sin\alpha\), we have: \(\cos\alpha = 2\sin\beta\) Since the max value of cosine is 1 and the max value of sine is also 1, this relationship is valid. Choice B is true.

We'll start by finding expressions for the time of flights: Projectile 1: \(t_1 = \frac{2u\sin\alpha}{g}\) Projectile 2: \(t_2= \frac{2u\sin\beta}{g}\) The given relationship is: \(\frac{t_1}{t_2} = \tan\alpha\) Substituting our expressions for \(t_1\) and \(t_2\), we have: \(\frac{\frac{2u\sin\alpha}{g}}{\frac{2u\sin\beta}{g}} = \tan\alpha\) Simplifying, we get: \(\frac{\sin\alpha}{\sin\beta} = \tan\alpha\) Using the identity \(\tan x = \frac{\sin x}{\cos x}\), we can write: \(\frac{\sin\alpha}{\sin\beta} = \frac{\sin\alpha}{\cos\alpha}\) Equating the denominators, we get: \(\sin\beta = \cos\alpha\) We had a similar result in Choice B, which was true. Therefore, Choice C is also true.

The given relationship is: \(\tan\alpha = \sqrt{\frac{h_{1}}{h_{2}}}\) Substituting our expressions for \(h_1\) and \(h_2\), we have: \(\tan\alpha = \sqrt{\frac{\frac{u^2 \sin^2 \alpha}{2g}}{\frac{u^2 \sin^2 \beta}{2g}}}\) Simplifying, we get: \(\tan\alpha = \frac{\sin\alpha}{\sin\beta}\) Again, this is the same result we derived in Choice C, which was true. Therefore, Choice D is also true. In summary, Choices B, C, and D are true, while Choice A is false.