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:

• A. \( \tan\theta \)
• B. \( \tan^{2}\theta \)
• C. \( \cot\theta \)
• D. \( \cot^{2}\theta \)

Hint:

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 \( \tan\theta \).

Answer 1

The Correct Option is A

Step 1: Use the equal range fact.
For one fixed launch speed, two different angles give the same horizontal range. These two angles always add up to $90^{\circ}$. So if one angle is $\theta$, the other is $90^{\circ}-\theta$. They are called complementary angles.
Step 2: Recall the time of flight formula.
The total time a projectile stays in the air is: \[ T = \frac{2u\sin\alpha}{g} \] where $\alpha$ is the launch angle, $u$ is the speed and $g$ is gravity.
Step 3: Write the time for the first angle.
For angle $\theta$: \[ T_1 = \frac{2u\sin\theta}{g} \]
Step 4: Write the time for the second angle.
For angle $90^{\circ}-\theta$, and using $\sin(90^{\circ}-\theta)=\cos\theta$: \[ T_2 = \frac{2u\sin(90^{\circ}-\theta)}{g} = \frac{2u\cos\theta}{g} \]
Step 5: Divide to get the ratio.
Now divide the two times. The common parts $2u$ and $g$ cancel: \[ \frac{T_1}{T_2} = \frac{\sin\theta}{\cos\theta} \]
Step 6: Simplify and conclude.
Since $\dfrac{\sin\theta}{\cos\theta}=\tan\theta$, the ratio is: \[ \boxed{\frac{T_1}{T_2} = \tan\theta} \]