Two objects A and B are horizontal at angles `45^(@)` and `60^(@)` respectively with the horizontal. It is found that both the objects attain same maximum height . If `u_(A) and u_(B)` are initial speeds of projection of objects A and B respectively , then `u_(A)//u_(B)` is

Options

A. `sqrt((5)/(2))`
B. `sqrt((3)/(2))`
C. `sqrt((7)/(2))`
D. `sqrt((5)/(3))`

Explanation

To solve the problem, we need to find the ratio of the initial speeds of projection \( u_A \) and \( u_B \) for two objects A and B, which are projected at angles of \( 45^\circ \) and \( 60^\circ \) respectively, and attain the same maximum height.

### Step-by-Step Solution:

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

2. **Setting Up the Heights**:
For object A (projected at \( 45^\circ \)):
\[ H_1 = \frac{u_A^2 \sin^2(45^\circ)}{2g} \]
For object B (projected at \( 60^\circ \)):
\[ H_2 = \frac{u_B^2 \sin^2(60^\circ)}{2g} \]

3. **Equating the Heights**:
Since both objects attain the same maximum height, we can set \( H_1 \) equal to \( H_2 \):
\[ \frac{u_A^2 \sin^2(45^\circ)}{2g} = \frac{u_B^2 \sin^2(60^\circ)}{2g} \]
The \( 2g \) cancels out from both sides:
\[ u_A^2 \sin^2(45^\circ) = u_B^2 \sin^2(60^\circ) \]

4. **Substituting Values of Sine**:
We know:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \sin^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ \sin^2(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

5. **Substituting into the Equation**:
Now substituting these values into our equation:
\[ u_A^2 \cdot \frac{1}{2} = u_B^2 \cdot \frac{3}{4} \]
Rearranging gives:
\[ \frac{u_A^2}{u_B^2} = \frac{3}{4} \cdot 2 = \frac{3}{2} \]

6. **Taking the Square Root**:
Taking the square root of both sides to find the ratio of the initial speeds:
\[ \frac{u_A}{u_B} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

### Final Result:
The ratio of the initial speeds \( u_A \) and \( u_B \) is:
\[ \frac{u_A}{u_B} = \frac{\sqrt{3}}{\sqrt{2}} \]