A TV tower of height h can broadcast program upto a distance ( given radius of earth R ):

Options

A. `Rh`
B. `2Rh`
C. `sqrt(2Rh)`
D. `sqrt(Rh).`

Answer

To solve the problem of determining the maximum distance a TV tower of height \( h \) can broadcast a program given the radius of the Earth \( R \), we can use the following steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Visualize the Earth as a sphere with radius \( R \).
- The TV tower is at height \( h \) above the Earth's surface.
- The maximum distance \( d \) that the signal can reach is along the line of sight from the top of the tower to the horizon.

2. **Use the Pythagorean Theorem**:
- Consider a right triangle formed by:
- The radius of the Earth \( R \) (one leg),
- The height of the tower \( h \) added to the radius of the Earth \( R + h \) (the other leg),
- The distance \( d \) from the tower to the horizon (the hypotenuse).
- According to the Pythagorean theorem:
\[ (R + h)^2 = R^2 + d^2 \]

3. **Expand the Equation**:
- Expanding the left side:
\[ R^2 + 2Rh + h^2 = R^2 + d^2 \]

4. **Rearranging the Equation**:
- Subtract \( R^2 \) from both sides:
\[ 2Rh + h^2 = d^2 \]

5. **Neglecting \( h^2 \)**:
- If \( h \) is much smaller than \( R \) (which is generally true for TV towers), we can neglect \( h^2 \):
\[ d^2 \approx 2Rh \]

6. **Taking the Square Root**:
- Taking the square root of both sides gives:
\[ d \approx \sqrt{2Rh} \]

7. **Final Result**:
- Therefore, the maximum distance \( d \) that the TV tower can broadcast is:
\[ d = \sqrt{2Rh} \]

### Conclusion:
The correct option for the maximum distance a TV tower of height \( h \) can broadcast, given the radius of the Earth \( R \), is:
\[ d = \sqrt{2Rh} \]