Two bodies are thrown up at angles of `45^(@)` and `60^(@)`, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

Options

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

Answer

To solve the problem, we need to find the ratio of the initial velocities of two bodies thrown at angles of \(45^\circ\) and \(60^\circ\) that reach the same maximum vertical height.

### Step-by-Step Solution:

1. **Understanding the 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 velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

2. **Setting up the equations for both bodies**:
For the first body thrown at \(45^\circ\):
\[ H_1 = \frac{u_1^2 \sin^2(45^\circ)}{2g} \]
For the second body thrown at \(60^\circ\):
\[ H_2 = \frac{u_2^2 \sin^2(60^\circ)}{2g} \]

3. **Since both bodies attain the same height**:
We can set \(H_1 = H_2\):
\[ \frac{u_1^2 \sin^2(45^\circ)}{2g} = \frac{u_2^2 \sin^2(60^\circ)}{2g} \]
The \(2g\) cancels out from both sides:
\[ u_1^2 \sin^2(45^\circ) = u_2^2 \sin^2(60^\circ) \]

4. **Substituting the values of sine**:
We know that:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
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:
\[ u_1^2 \cdot \frac{1}{2} = u_2^2 \cdot \frac{3}{4} \]

5. **Cross-multiplying to find the ratio**:
Rearranging the equation:
\[ 4u_1^2 = 6u_2^2 \]
Dividing both sides by \(u_2^2\):
\[ \frac{u_1^2}{u_2^2} = \frac{6}{4} = \frac{3}{2} \]

6. **Taking the square root to find the ratio of velocities**:
Taking the square root of both sides gives:
\[ \frac{u_1}{u_2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

7. **Final Ratio**:
Therefore, the ratio of the velocities \(u_1 : u_2\) is:
\[ u_1 : u_2 = \sqrt{3} : \sqrt{2} \]

### Conclusion:
The ratio of the velocities with which the two bodies are thrown is \(\sqrt{3} : \sqrt{2}\).