The velocity of a freely falling body changes as `g^ph^q`where g is acceleration due to gravity and h is the height. The values of p and q are

Options

A. `1`,`(1)/(2)`
B. `(1)/(2)`,`(1)/(2)`
C. `(1)/(2)`,`1`
D. 1,1

Explanation

To solve the problem, we need to determine the values of \( p \) and \( q \) in the equation \( v = k g^p h^q \), where \( v \) is the velocity of a freely falling body, \( g \) is the acceleration due to gravity, and \( h \) is the height.

### Step-by-step Solution:

1. **Identify the Dimensions**:
- The dimensional formula for velocity \( v \) is given by:
\[ [v] = L T^{-1} \]
- The dimensional formula for acceleration due to gravity \( g \) is:
\[ [g] = L T^{-2} \]
- The dimensional formula for height \( h \) is:
\[ [h] = L \]

2. **Write the Dimensional Equation**:
- The expression for velocity can be rewritten in terms of dimensions:
\[ [v] = k [g]^p [h]^q \]
- Substituting the dimensional formulas, we have:
\[ L T^{-1} = k (L T^{-2})^p (L)^q \]

3. **Expand the Right Side**:
- Expanding the right side gives:
\[ L T^{-1} = k L^p T^{-2p} L^q \]
- This simplifies to:
\[ L T^{-1} = k L^{p+q} T^{-2p} \]

4. **Equate Dimensions**:
- Since \( k \) is dimensionless, we can equate the dimensions on both sides:
- For length \( L \):
\[ 1 = p + q \quad \text{(1)} \]
- For time \( T \):
\[ -1 = -2p \quad \text{(2)} \]

5. **Solve the Equations**:
- From equation (2):
\[ -1 = -2p \implies p = \frac{1}{2} \]
- Substitute \( p \) into equation (1):
\[ 1 = \frac{1}{2} + q \implies q = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Final Values**:
- Therefore, the values of \( p \) and \( q \) are:
\[ p = \frac{1}{2}, \quad q = \frac{1}{2} \]

### Final Answer:
The values of \( p \) and \( q \) are \( \frac{1}{2} \) and \( \frac{1}{2} \), respectively.