A body is thrown vertically up with certain initial velocity, the potential and kinetic energies of the body are equal is thrown with double the velocity upwards, the ratio of potential and kinetic energies upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

Options

A. (a)`1:1`
B. (b)`1:4`
C. (c)`1:7`
D. (d)`1:8`

Answer

To solve the problem, we need to find the ratio of potential energy (U) to kinetic energy (K) when a body is thrown upwards with double the initial velocity, specifically at the same height where the potential and kinetic energies were equal for the first case.

### Step-by-Step Solution:

1. **Initial Condition with Velocity \( V_0 \)**:
- When the body is thrown upwards with an initial velocity \( V_0 \), at a certain height \( h \), the potential energy (U) and kinetic energy (K) are equal.
- Let’s denote the potential energy at height \( h \) as \( U_0 \) and the kinetic energy at that height as \( K_0 \).
- Since \( U_0 = K_0 \), we can express this as:
\[ U_0 = K_0 = \frac{1}{2} m V_0^2 \]
- The potential energy at height \( h \) can also be expressed as:
\[ U_0 = mgh \]
- Setting these equal gives:
\[ mgh = \frac{1}{2} m V_0^2 \]
- From this, we can find the height \( h \):
\[ h = \frac{V_0^2}{2g} \]

2. **Condition with Double Velocity \( 2V_0 \)**:
- Now, consider the case where the body is thrown upwards with an initial velocity \( 2V_0 \).
- The initial kinetic energy \( K_0' \) is:
\[ K_0' = \frac{1}{2} m (2V_0)^2 = 2m V_0^2 \]
- At the same height \( h \), the potential energy \( U_0 \) remains the same:
\[ U_0 = mgh = \frac{1}{2} m V_0^2 \]

3. **Finding Kinetic Energy at Height \( h \)**:
- The total mechanical energy is conserved. Therefore, the total energy at the initial point (when thrown with \( 2V_0 \)) is equal to the total energy at height \( h \):
\[ K_0' + U_0 = K + U \]
- At height \( h \):
\[ 2m V_0^2 + \frac{1}{2} m V_0^2 = K + \frac{1}{2} m V_0^2 \]
- Simplifying gives:
\[ 2.5 m V_0^2 = K + \frac{1}{2} m V_0^2 \]
- Rearranging to find \( K \):
\[ K = 2.5 m V_0^2 - \frac{1}{2} m V_0^2 = 2m V_0^2 \]

4. **Finding the Ratio of Potential Energy to Kinetic Energy**:
- We already know that:
\[ U = \frac{1}{2} m V_0^2 \]
- And we found:
\[ K = 2m V_0^2 \]
- The ratio \( \frac{U}{K} \) is:
\[ \frac{U}{K} = \frac{\frac{1}{2} m V_0^2}{2m V_0^2} = \frac{1/2}{2} = \frac{1}{4} \]

### Final Answer:
The ratio of potential energy to kinetic energy when the body is thrown with double the velocity and crosses the same point is \( \frac{1}{4} \).