A body is projected vertically up. At certain heighth above the ground, the ratio of its potential energy to the kinetic energy is 2:3. At what height above the ground, the ratio of its kinetic energy to the potential energy will be 2 : 3 ?

Options

A. `(2)/(5)h`
B. `(3)/(2)h`
C. `(4)/(7)h`
D. `(9)/(8)h`

Answer

To solve the problem step by step, we need to analyze the relationship between potential energy (PE) and kinetic energy (KE) at two different heights.

### Step 1: Understand the given ratio of PE to KE
At a certain height \( h_1 \), the ratio of potential energy to kinetic energy is given as:
\[ \frac{PE}{KE} = \frac{2}{3} \]

### Step 2: Express PE and KE in terms of height
Potential energy (PE) at height \( h_1 \) is given by:
\[ PE = mgh_1 \]
Kinetic energy (KE) at height \( h_1 \) can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Using the conservation of mechanical energy, we know that the total mechanical energy at the point of projection is equal to the total mechanical energy at height \( h_1 \):
\[ \text{Total Energy} = KE + PE = \frac{1}{2} mv_0^2 \]
where \( v_0 \) is the initial velocity when the body was projected.

### Step 3: Set up the equation based on the ratio
From the ratio \( \frac{PE}{KE} = \frac{2}{3} \), we can express this as:
\[ PE = \frac{2}{3} KE \]
Substituting the expressions for PE and KE, we have:
\[ mgh_1 = \frac{2}{3} \left(\frac{1}{2} mv^2\right) \]

### Step 4: Simplify the equation
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh_1 = \frac{1}{3} v^2 \]
This equation will help us find the height \( h_1 \) in terms of the velocity \( v \).

### Step 5: Find the height where the ratio of KE to PE is 2:3
Now, we need to find the height \( h_2 \) where the ratio of kinetic energy to potential energy is:
\[ \frac{KE}{PE} = \frac{2}{3} \]
This can be expressed as:
\[ KE = \frac{2}{3} PE \]
Substituting the expressions for PE and KE:
\[ \frac{1}{2} mv^2 = \frac{2}{3} (mgh_2) \]

### Step 6: Set up the equation based on the new ratio
Cancelling \( m \) from both sides:
\[ \frac{1}{2} v^2 = \frac{2}{3} gh_2 \]

### Step 7: Relate the two equations
From the previous step, we have:
\[ gh_1 = \frac{1}{3} v^2 \]
Substituting \( v^2 \) from this equation into the new equation:
\[ \frac{1}{2} \left(3gh_1\right) = \frac{2}{3} gh_2 \]
This simplifies to:
\[ \frac{3}{2} gh_1 = \frac{2}{3} gh_2 \]

### Step 8: Solve for \( h_2 \)
Cancelling \( g \) from both sides:
\[ \frac{3}{2} h_1 = \frac{2}{3} h_2 \]
Cross-multiplying gives:
\[ 3 \cdot 3 h_2 = 2 \cdot 2 h_1 \]
Thus,
\[ h_2 = \frac{4}{9} h_1 \]

### Step 9: Final height expression
Now, we can express \( h_2 \) in terms of \( h_1 \):
\[ h_2 = \frac{4}{9} h_1 \]

### Conclusion
The height above the ground where the ratio of kinetic energy to potential energy will be \( 2:3 \) is \( \frac{4}{9} h_1 \).