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Question: Two projectiles were launched from same position simultaneously with same speed. One of the projecti...

Two projectiles were launched from same position simultaneously with same speed. One of the projectile was launched at angle (45−α)∘(45-\alpha)^{\circ} and the other at an angle of (45+α)∘(45+\alpha)^{\circ} with the horizontal. Find the ratio of maximum height of the projectiles.

A

1−cos⁡2α1+cos⁡2α\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}

B

1−sin⁡2α1+sin⁡2α\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}

C

1−tan⁡α1+tan⁡α\frac{1-\tan \alpha}{1+\tan \alpha}

D

1−sin⁡α1+sin⁡α\frac{1-\sin \alpha}{1+\sin \alpha}

Answer

1−tan⁡α1+tan⁡α\frac{1-\tan \alpha}{1+\tan \alpha}

Explanation

Solution

The maximum height of a projectile is given by the formula H=u2sin⁡2θ2gH = \frac{u^{2} \sin^{2} \theta}{2g}. For the two projectiles, we have: H1=u2sin⁡2(45−α)2gH_{1} = \frac{u^{2} \sin^{2}(45-\alpha)}{2g} and H2=u2sin⁡2(45+α)2gH_{2} = \frac{u^{2} \sin^{2}(45+\alpha)}{2g}. Using the identity sin⁡(45±α)=22cos⁡α±22sin⁡α\sin(45 \pm \alpha) = \frac{\sqrt{2}}{2} \cos \alpha \pm \frac{\sqrt{2}}{2} \sin \alpha, we can find the heights and then take the ratio. After simplification, we find that the ratio of maximum heights is 1−tan⁡α1+tan⁡α\frac{1-\tan \alpha}{1+\tan \alpha}.