Two particles are projected with same velocity but at angles of projection 35° and 55°. Then their horizontal ranges are in the ratio of

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. 4:1`

Answer

To solve the problem of finding the ratio of the horizontal ranges of two particles projected at angles of 35° and 55° with the same initial velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Let the initial velocity of both particles be \( u \).
- The angles of projection are \( \theta_1 = 35^\circ \) and \( \theta_2 = 55^\circ \).

2. **Formula for Horizontal Range:**
- The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

3. **Calculate the Ranges for Each Angle:**
- For the first particle (angle \( \theta_1 \)):
\[ R_1 = \frac{u^2 \sin(2 \theta_1)}{g} = \frac{u^2 \sin(70^\circ)}{g} \]
- For the second particle (angle \( \theta_2 \)):
\[ R_2 = \frac{u^2 \sin(2 \theta_2)}{g} = \frac{u^2 \sin(110^\circ)}{g} \]

4. **Find the Ratio of the Ranges:**
- The ratio of the ranges \( R_1 \) and \( R_2 \) can be calculated as follows:
\[ \frac{R_1}{R_2} = \frac{\sin(70^\circ)}{\sin(110^\circ)} \]

5. **Use the Identity for Sine:**
- We know that \( \sin(110^\circ) = \sin(180^\circ - 70^\circ) = \sin(70^\circ) \).
- Therefore, we can simplify the ratio:
\[ \frac{R_1}{R_2} = \frac{\sin(70^\circ)}{\sin(70^\circ)} = 1 \]

6. **Conclusion:**
- The ratio of the horizontal ranges \( R_1 : R_2 = 1 : 1 \).

### Final Answer:
The horizontal ranges of the two particles are in the ratio \( 1 : 1 \).