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(2/3)`
B. `2/sqrt3`
C. `sqrt(3/2)`
D. `sqrt3/2`

Explanation

To solve the problem, we need to find the ratio of the velocities of two bodies thrown at angles of \(45^\circ\) and \(60^\circ\) that reach the same vertical height. We will use the formula for the maximum height attained by a projectile:

\[ H = \frac{U^2 \sin^2 \theta}{2g} \]

where \(H\) is the maximum height, \(U\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

### Step-by-step Solution:

1. **Define the Variables**:
Let \(U_1\) be the initial velocity of the body thrown at \(45^\circ\) and \(U_2\) be the initial velocity of the body thrown at \(60^\circ\).

2. **Write the Height Equations**:
For the body thrown at \(45^\circ\):
\[ H_1 = \frac{U_1^2 \sin^2(45^\circ)}{2g} \]
For the body thrown at \(60^\circ\):
\[ H_2 = \frac{U_2^2 \sin^2(60^\circ)}{2g} \]

3. **Substitute the Values of Sine**:
We know that:
\(\sin(45^\circ) = \frac{1}{\sqrt{2}}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
Therefore,
\[ H_1 = \frac{U_1^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{U_1^2 \cdot \frac{1}{2}}{2g} = \frac{U_1^2}{4g} \]
\[ H_2 = \frac{U_2^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{U_2^2 \cdot \frac{3}{4}}{2g} = \frac{3U_2^2}{8g} \]

4. **Set the Heights Equal**:
Since both bodies attain the same height:
\[ H_1 = H_2 \]
Therefore,
\[ \frac{U_1^2}{4g} = \frac{3U_2^2}{8g} \]

5. **Eliminate \(g\)**:
We can cancel \(g\) from both sides:
\[ \frac{U_1^2}{4} = \frac{3U_2^2}{8} \]

6. **Cross Multiply**:
\[ 8U_1^2 = 12U_2^2 \]

7. **Rearrange to Find the Ratio**:
\[ \frac{U_1^2}{U_2^2} = \frac{12}{8} = \frac{3}{2} \]
Taking the square root of both sides gives:
\[ \frac{U_1}{U_2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

### Final Answer:
The ratio of the velocities with which the two bodies are thrown is:
\[ \frac{U_1}{U_2} = \frac{\sqrt{3}}{\sqrt{2}} \]