Two balls are dropped from the same height at two different places A and B where the acceleration due to gravities are `g_A and g_B`. The body at 'B' takes 't' seconds less to reach the ground and strikes the ground with a velocity greater than at 'A' by `upsilon m//s`. Then the value of `upsilon//t` is

Options

A. `1/(sqrt(g_A g_B))`
B. `2sqrt(g_A g_B)`
C. `1/(g_A g_B)`
D. `sqrt(g_A g_B)`

Answer

To solve the problem, we need to analyze the motion of the two balls dropped from the same height at two different places A and B, where the acceleration due to gravity is different at each location.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the height from which the balls are dropped be \( h \).
- The acceleration due to gravity at A is \( g_A \) and at B is \( g_B \).
- Let \( t_A \) be the time taken by the ball at A to reach the ground and \( t_B \) be the time taken by the ball at B.
- According to the problem, \( t_A = t_B + t \) (where \( t \) is the time difference).

2. **Using the Equation of Motion**:
- For ball A, using the equation \( h = \frac{1}{2} g_A t_A^2 \):
\[ h = \frac{1}{2} g_A t_A^2 \implies t_A^2 = \frac{2h}{g_A} \implies t_A = \sqrt{\frac{2h}{g_A}} \]

- For ball B, using the equation \( h = \frac{1}{2} g_B t_B^2 \):
\[ h = \frac{1}{2} g_B t_B^2 \implies t_B^2 = \frac{2h}{g_B} \implies t_B = \sqrt{\frac{2h}{g_B}} \]

3. **Relating the Times**:
- Since \( t_A = t_B + t \), we can express this as:
\[ \sqrt{\frac{2h}{g_A}} = \sqrt{\frac{2h}{g_B}} + t \]

4. **Squaring Both Sides**:
- Squaring both sides gives:
\[ \frac{2h}{g_A} = \left(\sqrt{\frac{2h}{g_B}} + t\right)^2 \]
- Expanding the right side:
\[ \frac{2h}{g_A} = \frac{2h}{g_B} + 2t\sqrt{\frac{2h}{g_B}} + t^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{2h}{g_A} - \frac{2h}{g_B} = 2t\sqrt{\frac{2h}{g_B}} + t^2 \]
- Factoring out \( 2h \):
\[ 2h \left(\frac{1}{g_A} - \frac{1}{g_B}\right) = 2t\sqrt{\frac{2h}{g_B}} + t^2 \]

6. **Finding the Velocities**:
- The final velocity of the ball at A is:
\[ v_A = g_A t_A = g_A \sqrt{\frac{2h}{g_A}} = \sqrt{2hg_A} \]
- The final velocity of the ball at B is:
\[ v_B = g_B t_B = g_B \sqrt{\frac{2h}{g_B}} = \sqrt{2hg_B} \]
- Given that \( v_B = v_A + \upsilon \):
\[ \sqrt{2hg_B} = \sqrt{2hg_A} + \upsilon \]

7. **Finding \( \frac{\upsilon}{t} \)**:
- From the equations, we can derive:
\[ \upsilon = \sqrt{2hg_B} - \sqrt{2hg_A} \]
- Using the time difference \( t \) and the relationships derived, we can find:
\[ \frac{\upsilon}{t} = \sqrt{g_A g_B} \]

### Final Result:
Thus, the value of \( \frac{\upsilon}{t} \) is:
\[ \frac{\upsilon}{t} = \sqrt{g_A g_B} \]