A particle `A` is projected verically upwards. Another indentical particle `B` is projected at an angle of `45^(@)`. Both reach the same height. The ratio of the initial kinetic energy of `A` to that of `B` is `-`

Options

A. `1//4`
B. `1//3`
C. `1//2`
D. `1`

Answer

To solve the problem, we need to find the ratio of the initial kinetic energy of particle A (projected vertically upwards) to that of particle B (projected at an angle of 45 degrees), given that both reach the same maximum height.

### Step-by-step Solution:

1. **Define Variables:**
- Let the initial velocity of particle A be \( V_A \).
- Let the initial velocity of particle B be \( V_B \).
- Both particles reach the same maximum height \( H \).

2. **Use the Kinematic Equation for Maximum Height:**
- For particle A (projected vertically), the maximum height \( H \) is given by:
\[ H = \frac{V_A^2}{2g} \]
- For particle B (projected at an angle of 45 degrees), the vertical component of its velocity is \( V_B \sin(45^\circ) = \frac{V_B}{\sqrt{2}} \). Therefore, the maximum height for particle B is:
\[ H = \frac{\left(\frac{V_B}{\sqrt{2}}\right)^2}{2g} = \frac{V_B^2}{4g} \]

3. **Set the Maximum Heights Equal:**
- Since both particles reach the same height:
\[ \frac{V_A^2}{2g} = \frac{V_B^2}{4g} \]
- Cancel \( g \) from both sides:
\[ V_A^2 = \frac{V_B^2}{2} \]

4. **Find the Ratio of Initial Velocities:**
- Rearranging gives:
\[ \frac{V_A^2}{V_B^2} = \frac{1}{2} \]

5. **Calculate the Ratio of Kinetic Energies:**
- The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m V^2 \]
- Therefore, the ratio of the initial kinetic energies of A and B is:
\[ \frac{K_A}{K_B} = \frac{\frac{1}{2} m V_A^2}{\frac{1}{2} m V_B^2} = \frac{V_A^2}{V_B^2} \]
- Substituting the ratio we found:
\[ \frac{K_A}{K_B} = \frac{1}{2} \]

6. **Final Result:**
- The ratio of the initial kinetic energy of A to that of B is:
\[ \frac{K_A}{K_B} = \frac{1}{2} \]

### Conclusion:
The ratio of the initial kinetic energy of particle A to that of particle B is \( \frac{1}{2} \).

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