Question

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:

• A. \( \frac{1-\tan \alpha}{1+\tan \alpha} \)
• B. \( \frac{1-\sin 2\alpha}{1+\sin 2\alpha} \)
• C. \( \frac{1+\sin 2\alpha}{1-\sin 2\alpha} \)
• D. \( \frac{1+\sin \alpha}{1-\sin \alpha} \)

Hint:

When dealing with projectile motion, always use the formula for maximum height and apply the appropriate angle values to determine the desired quantities.

Answer 1

The Correct Option is B

The maximum height attained by a projectile is given by \( H = \frac{v_0^2 \sin^2 \theta}{2g} \), where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
For angles \( (45^\circ - \alpha) \) and \( (45^\circ + \alpha) \), the heights can be written as: \[ H_1 = \frac{v_0^2 \sin^2 (45^\circ - \alpha)}{2g}, \quad H_2 = \frac{v_0^2 \sin^2 (45^\circ + \alpha)}{2g} \] Using trigonometric identities and simplifying, the ratio of heights becomes: \[ \frac{H_1}{H_2} = \frac{1-\sin 2\alpha}{1+\sin 2\alpha} \]