Two projectiles are fired with the same initi | vedclass

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(C) The formula for the maximum height attained by a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given the initial speed $u$ is the same for both

Vedclass · Accepted Answer

$\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$

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(C) The formula for the maximum height attained by a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given the initial speed $u$ is the same for both projectiles,the ratio of their maximum heights is given by:$\frac{H_1}{H_2} = \frac{\sin^2(45^{\circ}-\alpha)}{\sin^2(45^{\circ}+\alpha)}$.Using the trigonometric expansion $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$\sin(45^{\circ}-\alpha) = \frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha = \frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)$.$\sin(45^{\circ}+\alpha) = \frac{1}{\sqrt{2}}\cos \alpha + \frac{1}{\sqrt{2}}\sin \alpha = \frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha)$.Squaring these expressions:$\sin^2(45^{\circ}-\alpha) = \frac{1}{2}(\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha) = \frac{1}{2}(1 - \sin 2\alpha)$.$\sin^2(45^{\circ}+\alpha) = \frac{1}{2}(\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) = \frac{1}{2}(1 + \sin 2\alpha)$.Therefore,the ratio is $\frac{H_1}{H_2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$.

Two projectiles are fired with the same initial speed from the same point on the ground at angles of $(45^{\circ}-\alpha)$ and $(45^{\circ}+\alpha)$,respectively,with the horizontal direction. The ratio of their maximum heights attained is:

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