The acceleration of gravity can be measured by projecting a body upward and measuring the time it takes to pass two given points in both directions.Show that if the time the body takes to pass a horizontal line `a` in both directions is `t_(A)` anytime to go by a second line `B` in both direction is `t_(B)`, then assuming that the acceleration is constant, its magnitude is `g=` (where `h` is the height of the line `B` above line `A`.)

Options

A. `h/(t_(A)^(2)-t_(B)^(2))`
B. `8h/(t_(A)^(2)-t_(B)^(2))`
C. `8h/(t_(A)^(2)+t_(B)^(2))`
D. `4h/(t_(A)^(2)+t_(B)^(2))`

Explanation

To solve the problem, we need to analyze the motion of a body projected upward under the influence of gravity. We will use the equations of motion to derive the expression for the acceleration due to gravity \( g \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
When a body is projected upwards, it moves against the gravitational force until it reaches its maximum height, after which it falls back down. We will consider two horizontal lines: line A (at height \( h_A \)) and line B (at height \( h_B \)), where \( h_B \) is above \( h_A \) by a height \( h \).

2. **Time to Pass Line A**:
Let \( t_A \) be the time taken to pass line A in both directions (upward and downward). The total time for the upward and downward journey through line A is \( t_A \).

3. **Time to Pass Line B**:
Let \( t_B \) be the time taken to pass line B in both directions. Similarly, the total time for the upward and downward journey through line B is \( t_B \).

4. **Using the Equations of Motion**:
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration (which is \(-g\) when going upwards), and \( t \) is the time.

5. **For Line A**:
When the body passes line A:
- Upward:
\[ h_A = ut_A - \frac{1}{2} g t_A^2 \]
- Downward:
\[ h_A = u(t_A + t_{up}) - \frac{1}{2} g (t_A + t_{up})^2 \]
Here, \( t_{up} \) is the time taken to reach the maximum height from line A.

6. **For Line B**:
When the body passes line B:
- Upward:
\[ h_B = ut_B - \frac{1}{2} g t_B^2 \]
- Downward:
\[ h_B = u(t_B + t_{down}) - \frac{1}{2} g (t_B + t_{down})^2 \]
Here, \( t_{down} \) is the time taken to fall back down from line B.

7. **Relating the Times**:
We can relate the times \( t_A \) and \( t_B \) to the height difference \( h \):
\[ h = h_B - h_A \]
By substituting the expressions for \( h_A \) and \( h_B \) into this equation, we can derive a relationship involving \( g \).

8. **Final Expression**:
After manipulating the equations and eliminating the initial velocity \( u \), we arrive at:
\[ g = \frac{2h}{t_B^2 - t_A^2} \]

### Conclusion:
Thus, the magnitude of the acceleration due to gravity \( g \) can be expressed as:
\[ g = \frac{2h}{t_B^2 - t_A^2} \]