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

To determine the ratio of maximum heights achieved by two projectiles, we employ the formula for projectile maximum height: \(H = \frac{v_0^2 \sin^2 \theta}{2g}\), where \(v_0\) is initial velocity, \(\theta\) is projection angle, and \(g\) is gravitational acceleration.

Given projection angles of \( (45^\circ - \alpha) \) and \( (45^\circ + \alpha) \), the corresponding maximum heights, \(H_1\) and \(H_2\), are calculated as:

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 these maximum heights is:

\(\frac{H_1}{H_2} = \frac{\sin^2 (45^\circ - \alpha)}{\sin^2 (45^\circ + \alpha)}\)

Applying the trigonometric identity \(\sin (A \pm B) = \sin A \cos B \pm \cos A \sin B\):

\(\sin (45^\circ - \alpha) = \sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha = \frac{\sqrt{2}}{2}(\cos \alpha - \sin \alpha)\)

\(\sin (45^\circ + \alpha) = \sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha = \frac{\sqrt{2}}{2}(\cos \alpha + \sin \alpha)\)

Substituting these into the ratio yields:

\(\frac{H_1}{H_2} = \left(\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2\)

Dividing the numerator and denominator by \((\cos \alpha)^2\):

\(\frac{1 - \frac{2\sin \alpha \cos \alpha}{1+\sin^2 \alpha}}{1 + \frac{2\sin \alpha \cos \alpha}{1+\sin^2 \alpha}}\)

Utilizing the double angle identity \(\sin 2\alpha = 2\sin \alpha \cos \alpha\), the expression simplifies to:

\(\frac{1-\sin 2\alpha}{1+\sin 2\alpha}\)

Therefore, the ratio of the maximum heights is \( \frac{1-\sin 2\alpha}{1+\sin 2\alpha} \).