Question

Two projectiles are fired from the ground with the same initial speeds from the same point at angles $ (45^\circ + \alpha) $ and $ (45^\circ - \alpha) $ with the horizontal direction. The ratio of their times of flights is:

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

Hint:

When dealing with projectile motion at different angles, the ratio of their times of flight can be found using trigonometric identities and the general equation for the time of flight of a projectile.

Answer 1

The Correct Option is D

Two projectiles are launched with identical initial speeds at angles \( (45^\circ + \alpha) \) and \( (45^\circ - \alpha) \) relative to the horizontal. The formula for the time of flight \( T \) of a projectile is \( T = \frac{2u \sin \theta}{g} \), where \( u \) is the initial speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.
Step 1: Time of flight for the first projectile
The first projectile is launched at an angle \( (45^\circ + \alpha) \). Its time of flight \( T_1 \) is calculated as:
\[ T_1 = \frac{2u \sin(45^\circ + \alpha)}{g} \]
Applying the sine angle addition identity:
\[ T_1 = \frac{2u (\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha)}{g} \]
Substituting \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ T_1 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha) \right)}{g} \]
Step 2: Time of flight for the second projectile
The second projectile is launched at an angle \( (45^\circ - \alpha) \). Its time of flight \( T_2 \) is:
\[ T_2 = \frac{2u \sin(45^\circ - \alpha)}{g} \]
Applying the sine angle subtraction identity:
\[ T_2 = \frac{2u (\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha)}{g} \]
Substituting \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ T_2 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha) \right)}{g} \] 
Step 3: Ratio of the times of flight
The ratio of the times of flight \( T_1 \) to \( T_2 \) is:
\[ \frac{T_1}{T_2} = \frac{\frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha) \right)}{g}}{\frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha) \right)}{g}} \]
Simplifying the expression yields:
\[ \frac{T_1}{T_2} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \]
Dividing the numerator and denominator by \( \cos \alpha \) and using the identity \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), the ratio becomes:
\[ \frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha} \] 
The correct answer is option (4): \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \).