Two particles are projected from the same point with the same speed at different angles `theta_1` & `theta_2` to the horizontal. They have the same range. Their times of flight are `t_1` & `t_2` respectily

Options

A. `(t_1)/(t_2)=tan^2theta_1`
B. `(t_1)/(sintheta_1)=(t_2)/(costheta_2)`
C. `(t_1)/(t_2)=tantheta_1`
D. `(t_1)/(t_2)=tan^2theta_2`

Explanation

To solve the problem step by step, we will analyze the given information about the two particles projected at different angles but with the same speed and the same range.

### Step 1: Understand the Range Formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Set Up the Condition for Equal Ranges
Since both particles have the same range, we can write:
\[ R_1 = R_2 \]
This implies:
\[ \frac{u^2 \sin 2\theta_1}{g} = \frac{u^2 \sin 2\theta_2}{g} \]
Cancelling out common terms, we have:
\[ \sin 2\theta_1 = \sin 2\theta_2 \]

### Step 3: Use the Property of Complementary Angles
From trigonometric identities, if \( \sin 2\theta_1 = \sin 2\theta_2 \), then:
\[ 2\theta_1 + 2\theta_2 = 180^\circ \]
This leads to:
\[ \theta_1 + \theta_2 = 90^\circ \]
This means that the angles of projection are complementary.

### Step 4: Write the Time of Flight Equations
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin \theta}{g} \]
Thus, for the two particles, we have:
\[ t_1 = \frac{2u \sin \theta_1}{g} \]
\[ t_2 = \frac{2u \sin \theta_2}{g} \]

### Step 5: Substitute \( \theta_2 \)
Since \( \theta_2 = 90^\circ - \theta_1 \), we can express \( t_2 \) as:
\[ t_2 = \frac{2u \sin(90^\circ - \theta_1)}{g} \]
Using the identity \( \sin(90^\circ - \theta) = \cos \theta \), we get:
\[ t_2 = \frac{2u \cos \theta_1}{g} \]

### Step 6: Form the Ratio of Times of Flight
Now, we can form the ratio of the times of flight:
\[ \frac{t_1}{t_2} = \frac{\frac{2u \sin \theta_1}{g}}{\frac{2u \cos \theta_1}{g}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sin \theta_1}{\cos \theta_1} \]
Thus, we find:
\[ \frac{t_1}{t_2} = \tan \theta_1 \]

### Conclusion
The final result shows that the ratio of the times of flight of the two particles is related to the tangent of the angle of projection of the first particle.