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. `1 : 1`
B. `1 : 4`
C. `1 : 7`
D. `1 : 8`

Explanation

To solve the problem, we need to analyze the potential and kinetic energies of a body thrown vertically upwards with two different initial velocities.

### Step-by-step Solution:

1. **Identify Initial Conditions:**
- Let the initial velocity of the first body be \( u \).
- The mass of the body is \( m \).

2. **Energy at Point P for the First Body:**
- When the body is thrown with velocity \( u \), at the point P where potential energy (PE) equals kinetic energy (KE), we can express the energies as:
- Kinetic Energy (KE) = \( \frac{1}{2} m u^2 \)
- Potential Energy (PE) = \( mgh \) (where \( h \) is the height at point P)
- Since PE = KE at point P, we have:
\[ mgh = \frac{1}{2} m u^2 \]
- This implies:
\[ gh = \frac{1}{2} u^2 \quad \text{(1)} \]

3. **Energy at Point P for the Second Body:**
- Now consider the second body thrown with double the initial velocity, \( 2u \).
- The kinetic energy at the ground level (just before being thrown) is:
\[ KE' = \frac{1}{2} m (2u)^2 = 2mu^2 \]
- The total mechanical energy (E) when thrown is:
\[ E = KE' + PE = 2mu^2 + 0 = 2mu^2 \]
- At point P, the potential energy is still \( mgh \), and we need to find the kinetic energy at this point:
\[ E = KE' + PE' \quad \Rightarrow \quad 2mu^2 = KE' + mgh \]
- Rearranging gives:
\[ KE' = 2mu^2 - mgh \quad \text{(2)} \]

4. **Substituting from Equation (1):**
- From equation (1), we know \( gh = \frac{1}{2} u^2 \):
\[ KE' = 2mu^2 - m \left(\frac{1}{2} u^2\right) = 2mu^2 - \frac{1}{2} mu^2 = \frac{3}{2} mu^2 \]

5. **Finding the Ratio of Potential Energy to Kinetic Energy:**
- At point P for the second body:
- Potential Energy \( PE' = mgh = m \left(\frac{1}{2} u^2\right) = \frac{1}{2} mu^2 \)
- Kinetic Energy \( KE' = \frac{3}{2} mu^2 \)
- The ratio of potential energy to kinetic energy is:
\[ \frac{PE'}{KE'} = \frac{\frac{1}{2} mu^2}{\frac{3}{2} mu^2} = \frac{1}{3} \]

### Final Answer:
The ratio of potential energy to kinetic energy when the second body crosses the same point P is \( \frac{1}{3} \).