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

We are given two projectiles that are fired at angles \( (45^\circ + \alpha) \) and \( (45^\circ - \alpha) \) with the horizontal direction. Both projectiles have the same initial speed. 
The time of flight \( T \) for a projectile is given by the formula: \[ T = \frac{2u \sin \theta}{g} \] where: \( u \) is the initial speed, \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity. 
Step 1: Time of flight for the first projectile
For the first projectile, the angle of projection is \( (45^\circ + \alpha) \), so the time of flight \( T_1 \) is: \[ T_1 = \frac{2u \sin(45^\circ + \alpha)}{g} \] Using the angle addition identity for sine: \[ T_1 = \frac{2u (\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha)}{g} \] \[ T_1 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha) \right)}{g} \] 
Step 2: Time of flight for the second projectile
For the second projectile, the angle of projection is \( (45^\circ - \alpha) \), so the time of flight \( T_2 \) is: \[ T_2 = \frac{2u \sin(45^\circ - \alpha)}{g} \] Using the angle subtraction identity for sine: \[ T_2 = \frac{2u (\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha)}{g} \] \[ T_2 = \frac{2u \left(\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha) \right)}{g} \] 
Step 3: Ratio of the times of flight
The ratio of the times of flight \( T_1 \) and \( T_2 \) 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: \[ \frac{T_1}{T_2} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \] Using the identity \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \), we find that the ratio of the times of flight is: \[ \frac{1 + \tan \alpha}{1 - \tan \alpha} \] 
Thus, the correct answer is option (4): \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \).

Answer 2

Given: \[ \theta_1 = 45^\circ + \alpha, \quad \theta_2 = 45^\circ - \alpha \] The time of flight \( T \) is given by the formula: \[ T = \frac{2v \sin \theta}{g} \] Now, we compute the ratio of the times of flight \( T_1 \) and \( T_2 \): \[ \frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} \] Using the trigonometric identity for sine: \[ \frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha}{\frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha} \] Simplifying further: \[ \frac{T_1}{T_2} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \] Using the standard identity for tangent: \[ \frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha} \] \[ \boxed{\frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha}} \]