The velocity of a body which has fallen under gravity varies as `g^(a) h^(b)` where g is the acceleration due to gravity at a place and h is the height through which the body has fallen, a and b are given by

Options

A. a = 1, b = 1/2
B. a = b = 1
C. a = 1/2, b = 1
D. a = 1/2, b = 1/2

Answer

To solve the problem, we need to determine the values of \( a \) and \( b \) in the equation that describes the velocity \( v \) of a body falling under gravity, which varies as \( v \propto g^a h^b \). Here, \( g \) is the acceleration due to gravity and \( h \) is the height fallen.

### Step-by-Step Solution:

1. **Write the relationship**:
\[ v = k \cdot g^a \cdot h^b \]
where \( k \) is a constant.

2. **Identify the dimensions**:
- The dimension of velocity \( v \) is \( [L T^{-1}] \).
- The dimension of acceleration due to gravity \( g \) is \( [L T^{-2}] \).
- The dimension of height \( h \) is \( [L] \).

3. **Express the dimensions in terms of \( a \) and \( b \)**:
\[ [v] = [g^a] \cdot [h^b] \]
This translates to:
\[ [L T^{-1}] = ([L T^{-2}])^a \cdot ([L])^b \]

4. **Expand the dimensions**:
\[ [L T^{-1}] = [L^a T^{-2a}] \cdot [L^b] \]
Combining the dimensions gives:
\[ [L T^{-1}] = [L^{a+b} T^{-2a}] \]

5. **Set up equations by comparing dimensions**:
From the left side and right side, we can equate the powers of \( L \) and \( T \):
- For \( L \):
\[ a + b = 1 \quad \text{(1)} \]
- For \( T \):
\[ -2a = -1 \quad \text{(2)} \]

6. **Solve the equations**:
From equation (2):
\[ -2a = -1 \implies a = \frac{1}{2} \]

Substitute \( a \) into equation (1):
\[ \frac{1}{2} + b = 1 \implies b = 1 - \frac{1}{2} = \frac{1}{2} \]

7. **Final values**:
Thus, we find:
\[ a = \frac{1}{2}, \quad b = \frac{1}{2} \]

### Conclusion:
The values of \( a \) and \( b \) are both \( \frac{1}{2} \).