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:
Show Hint
- \( \frac{1-\tan \alpha}{1+\tan \alpha} \)
- \( \frac{1-\sin 2\alpha}{1+\sin 2\alpha} \)
- \( \frac{1+\sin 2\alpha}{1-\sin 2\alpha} \)
- \( \frac{1+\sin \alpha}{1-\sin \alpha} \)
The Correct Option is B
Solution and Explanation
To find the ratio of the maximum heights attained by the two projectiles, we start with the formula for maximum height reached by a projectile: H = \( \frac{v_0^2 \sin^2 \theta}{2g} \) where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
The two angles are \( (45^\circ - \alpha) \) and \( (45^\circ + \alpha) \). Therefore, the maximum heights H1 and H2 for the two angles are:
H1 = \( \frac{v_0^2 \sin^2 (45^\circ - \alpha)}{2g} \)
H2 = \( \frac{v_0^2 \sin^2 (45^\circ + \alpha)}{2g} \)
The ratio of the maximum heights is:
\(\frac{H_1}{H_2} = \frac{\sin^2 (45^\circ - \alpha)}{\sin^2 (45^\circ + \alpha)}\)
Using the trigonometric identity \(\sin (A \pm B) = \sin A \cos B \pm \cos A \sin B\), we have:
\(\sin (45^\circ - \alpha) = \sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha = \frac{\sqrt{2}}{2}(\cos \alpha - \sin \alpha)\)
\(\sin (45^\circ + \alpha) = \sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha = \frac{\sqrt{2}}{2}(\cos \alpha + \sin \alpha)\)
Thus,
\(\frac{H_1}{H_2} = \left(\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2\)
By multiplying numerator and denominator by \(\frac{1}{(\cos \alpha)^2}\), we have:
\(\frac{1 - \frac{2\sin \alpha \cos \alpha}{1+\sin^2 \alpha}}{1 + \frac{2\sin \alpha \cos \alpha}{1+\sin^2 \alpha}}\)
Using \(\sin 2\alpha = 2\sin \alpha \cos \alpha\), the expression becomes:
\(\frac{1-\sin 2\alpha}{1+\sin 2\alpha}\)
Thus, the ratio of the maximum heights is \( \frac{1-\sin 2\alpha}{1+\sin 2\alpha} \).
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