Two bodies are projected at angle ` theta` and (`90 -theta`) to the horizontal with the same speed. Find the ration of their time of flight.

Options

A. `1: tan theta `
B. ` tan ^(2) theta :1`
C. ` 1:1`
D. `tan theta :1`

Answer

To find the ratio of the time of flight for two bodies projected at angles \( \theta \) and \( 90 - \theta \) to the horizontal with the same speed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time of Flight Formula**:
The time of flight \( T \) for a projectile launched with an initial velocity \( u \) at an angle \( \alpha \) to the horizontal is given by:
\[ T = \frac{2u \sin \alpha}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Calculate Time of Flight for the First Body**:
For the first body projected at an angle \( \theta \):
\[ T_1 = \frac{2u \sin \theta}{g} \]

3. **Calculate Time of Flight for the Second Body**:
For the second body projected at an angle \( 90 - \theta \):
\[ T_2 = \frac{2u \sin(90 - \theta)}{g} \]
Using the trigonometric identity \( \sin(90 - \theta) = \cos \theta \):
\[ T_2 = \frac{2u \cos \theta}{g} \]

4. **Find the Ratio of Time of Flights**:
To find the ratio \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \frac{\frac{2u \sin \theta}{g}}{\frac{2u \cos \theta}{g}} \]
The \( 2u \) and \( g \) terms cancel out:
\[ \frac{T_1}{T_2} = \frac{\sin \theta}{\cos \theta} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \tan \theta \]

5. **Express the Ratio**:
Thus, the ratio of the time of flight of the two bodies is:
\[ T_1 : T_2 = \tan \theta : 1 \]

### Final Answer:
The ratio of their time of flight is \( \tan \theta : 1 \).