Two balls are projected making angles of 30° and 45°respectively with the horizontal. If both have same velocity at the highest point of their path, then the ratio of their horizontal ranges is 

Options

A. 1:3
B. `3:1`
C. `sqrt 3 : sqrt2`
D. `1 :sqrt3`

Explanation

To solve the problem, we need to find the ratio of the horizontal ranges of two balls projected at angles of 30° and 45° with the same velocity at their highest point.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two balls projected at angles \( \theta_1 = 30^\circ \) and \( \theta_2 = 45^\circ \).
- Both balls have the same velocity at the highest point of their trajectory.

2. **Velocity at the Highest Point**:
- At the highest point of projectile motion, the vertical component of the velocity becomes zero. Thus, the velocity at the highest point is entirely horizontal.
- The horizontal component of the velocity for both balls can be expressed as:
\[ v_{1x} = v_1 \cos(30^\circ) \quad \text{and} \quad v_{2x} = v_2 \cos(45^\circ) \]
- Since both velocities at the highest point are equal, we have:
\[ v_1 \cos(30^\circ) = v_2 \cos(45^\circ) \]

3. **Calculating the Cosine Values**:
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)

4. **Setting Up the Equation**:
- Substituting the cosine values into the equation gives:
\[ v_1 \cdot \frac{\sqrt{3}}{2} = v_2 \cdot \frac{1}{\sqrt{2}} \]
- Rearranging the equation to find \( v_1 \) in terms of \( v_2 \):
\[ v_1 = v_2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{3}} = v_2 \cdot \frac{2\sqrt{2}}{3} \]

5. **Finding the Horizontal Ranges**:
- The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- For the first ball (30°):
\[ R_1 = \frac{v_1^2 \sin(60^\circ)}{g} = \frac{v_1^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
- For the second ball (45°):
\[ R_2 = \frac{v_2^2 \sin(90^\circ)}{g} = \frac{v_2^2 \cdot 1}{g} \]

6. **Substituting \( v_1 \)**:
- Substitute \( v_1 = v_2 \cdot \frac{2\sqrt{2}}{3} \) into \( R_1 \):
\[ R_1 = \frac{\left(v_2 \cdot \frac{2\sqrt{2}}{3}\right)^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{v_2^2 \cdot \frac{4 \cdot 2}{9} \cdot \frac{\sqrt{3}}{2}}{g} = \frac{v_2^2 \cdot \frac{4\sqrt{3}}{9}}{g} \]

7. **Finding the Ratio of Ranges**:
- Now we can find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{\frac{v_2^2 \cdot \frac{4\sqrt{3}}{9}}{g}}{\frac{v_2^2}{g}} = \frac{4\sqrt{3}}{9} \]

8. **Final Result**:
- The ratio of the horizontal ranges \( R_1 : R_2 \) is:
\[ R_1 : R_2 = \frac{4\sqrt{3}}{9} : 1 \]