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:
Show Hint
- \( 1 \)
- \( \frac{1 - \tan \alpha}{1 + \tan \alpha} \)
- \( \frac{1 + \sin 2\alpha}{1 - \sin 2\alpha} \)
- \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \)
The Correct Option is D
Approach Solution - 1
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): \( \frac{1 + \tan \alpha}{1 - \tan \alpha} \).
Approach Solution -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: \[ \frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha} \] \[ \boxed{\frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha}} \]
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