Question:
Two bodies projected with same velocity at different angles attain same range. If the time of flights of the bodies are \( T_{1} \) and \( T_{2} \) respectively, then \( \frac{T_{1}}{T_{2}} \) is:
Two bodies projected with same velocity at different angles attain same range. If the time of flights of the bodies are \( T_{1} \) and \( T_{2} \) respectively, then \( \frac{T_{1}}{T_{2}} \) is:
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Complementary projectile pairings reveal beautiful structural relations. For instance, the product of their flight times directly tracks the range: \( T_1T_2 = \frac{2R}{g} \), while their ratio maps to \( \tan\theta \).
Updated On: Jun 8, 2026
- \( \tan\theta \)
- \( \tan^{2}\theta \)
- \( \cot\theta \)
- \( \cot^{2}\theta \)
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The Correct Option is A
Solution and Explanation
Concept:
For a given initial launching speed magnitude, a projectile path yields an identical horizontal range across exactly two complementary launch angle variations: \( \theta \) and \( (90^{\circ} - \theta) \). The corresponding general formula for total time of flight is \( T = \frac{2u\sin\alpha}{g} \).
Step 1: Formulating independent equations for the complementary flight durations.
Let the projection angles be \( \alpha_1 = \theta \) and \( \alpha_2 = 90^{\circ} - \theta \): \[ T_1 = \frac{2u\sin\theta}{g} \] \[ T_2 = \frac{2u\sin(90^{\circ} - \theta)}{g} = \frac{2u\cos\theta}{g} \]
Step 2: Evaluating the ratio.
Dividing \( T_1 \) by \( T_2 \): \[ \frac{T_1}{T_2} = \frac{\frac{2u\sin\theta}{g}}{\frac{2u\cos\theta}{g}} = \frac{\sin\theta}{\cos\theta} = \tan\theta \]
Step 1: Formulating independent equations for the complementary flight durations.
Let the projection angles be \( \alpha_1 = \theta \) and \( \alpha_2 = 90^{\circ} - \theta \): \[ T_1 = \frac{2u\sin\theta}{g} \] \[ T_2 = \frac{2u\sin(90^{\circ} - \theta)}{g} = \frac{2u\cos\theta}{g} \]
Step 2: Evaluating the ratio.
Dividing \( T_1 \) by \( T_2 \): \[ \frac{T_1}{T_2} = \frac{\frac{2u\sin\theta}{g}}{\frac{2u\cos\theta}{g}} = \frac{\sin\theta}{\cos\theta} = \tan\theta \]
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