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 comparing the maximum heights of projectiles, use the relationship between the initial velocity, launch angle, and the sine of the angle. This simplifies the calculation of the ratio of maximum heights.

Answer 1

The Correct Option is B

The maximum height \( H \) attained by a projectile is given by the formula: \[ H = \frac{v^2 \sin^2 \theta}{2g} \] where \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. For the two projectiles, let \( H_1 \) and \( H_2 \) be the maximum heights for the angles \( 45^\circ - \alpha \) and \( 45^\circ + \alpha \), respectively. The ratio of the maximum heights is: \[ \frac{H_1}{H_2} = \frac{\frac{v^2 \sin^2(45^\circ - \alpha)}{2g}}{\frac{v^2 \sin^2(45^\circ + \alpha)}{2g}} = \frac{\sin^2(45^\circ - \alpha)}{\sin^2(45^\circ + \alpha)} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha} \] Thus, the correct answer is \( \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha} \).