If 2 balls are projected at angles `45^(@)` and `60^(@)` and the maximum heights reached are same, what is the ratio of their initial velocities ?

Options

A. `sqrt(2) : sqrt(3)`
B. `sqrt(3):sqrt(2)`
C. `3:2`
D. `2:3`

Explanation

To solve the problem, we need to find the ratio of the initial velocities of two balls projected at angles of 45° and 60°, given that they reach the same maximum height.

### Step-by-Step Solution:

1. **Understanding the Maximum Height Formula**:
The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{U^2 \sin^2 \theta}{2g} \]
where \( U \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Setting Up the Equations for Both Balls**:
For the first ball projected at an angle of 45°:
\[ H_1 = \frac{U_1^2 \sin^2(45^\circ)}{2g} \]
For the second ball projected at an angle of 60°:
\[ H_2 = \frac{U_2^2 \sin^2(60^\circ)}{2g} \]

3. **Equating the Heights**:
Since the maximum heights are the same, we can set \( H_1 = H_2 \):
\[ \frac{U_1^2 \sin^2(45^\circ)}{2g} = \frac{U_2^2 \sin^2(60^\circ)}{2g} \]

4. **Canceling Common Terms**:
The \( 2g \) in the denominator cancels out:
\[ U_1^2 \sin^2(45^\circ) = U_2^2 \sin^2(60^\circ) \]

5. **Substituting the Sine Values**:
We know:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Substituting these values into the equation gives:
\[ U_1^2 \left(\frac{1}{\sqrt{2}}\right)^2 = U_2^2 \left(\frac{\sqrt{3}}{2}\right)^2 \]
Simplifying this:
\[ U_1^2 \cdot \frac{1}{2} = U_2^2 \cdot \frac{3}{4} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{U_1^2}{U_2^2} = \frac{3/4}{1/2} = \frac{3}{2} \]

7. **Taking the Square Root**:
Taking the square root of both sides to find the ratio of the initial velocities:
\[ \frac{U_1}{U_2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

### Final Result:
The ratio of their initial velocities \( U_1 : U_2 \) is:
\[ \sqrt{3} : \sqrt{2} \]