A stone projected vertically upwards from the ground reaches a maximum height `h`. When it is at a height `(3h)//(4)`, the ratio of its kinetic and potential energies is

Options

A. `3:1`
B. `1:1`
C. `1:3`
D. `1:2`

Answer

To solve the problem, we need to find the ratio of kinetic energy (KE) to potential energy (PE) when the stone is at a height of \( \frac{3h}{4} \).

### Step-by-Step Solution:

1. **Identify the maximum height and initial velocity**:
- The stone reaches a maximum height \( h \) when projected vertically upwards. At this height, the final velocity is zero.

2. **Use the kinematic equation to find initial velocity**:
- Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( v = 0 \) (at maximum height), \( a = -g \) (acceleration due to gravity), and \( s = h \).
\[ 0 = u^2 - 2gh \implies u^2 = 2gh \]

3. **Find the velocity at height \( \frac{3h}{4} \)**:
- Now, we need to find the velocity \( v \) when the stone is at height \( \frac{3h}{4} \). Again using the kinematic equation:
\[ v^2 = u^2 + 2as \]
Here, \( s = \frac{3h}{4} \):
\[ v^2 = 2gh - 2g \left( \frac{3h}{4} \right) \]
Simplifying this:
\[ v^2 = 2gh - \frac{3gh}{2} = \frac{4gh}{2} - \frac{3gh}{2} = \frac{gh}{2} \]

4. **Calculate the kinetic energy (KE) at height \( \frac{3h}{4} \)**:
- The kinetic energy is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting \( v^2 = \frac{gh}{2} \):
\[ KE = \frac{1}{2} m \left( \frac{gh}{2} \right) = \frac{mgh}{4} \]

5. **Calculate the potential energy (PE) at height \( \frac{3h}{4} \)**:
- The potential energy is given by:
\[ PE = mgh' \]
where \( h' = \frac{3h}{4} \):
\[ PE = mg \left( \frac{3h}{4} \right) = \frac{3mgh}{4} \]

6. **Find the ratio of kinetic energy to potential energy**:
- Now, we can find the ratio:
\[ \frac{KE}{PE} = \frac{\frac{mgh}{4}}{\frac{3mgh}{4}} = \frac{1}{3} \]

### Final Answer:
The ratio of kinetic energy to potential energy when the stone is at a height of \( \frac{3h}{4} \) is \( \frac{1}{3} \).

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