Two identical antennas mounted on identical towers are separated from each other by a distance of 45 km. What should nearly be the minimum height of receiving antenna to receive the signals in line of sight? (Assume radius of earth is 6400 km)

We need to find the minimum height of the receiving antenna $$h_r$$ to receive line-of-sight signals from a transmitting antenna of the same height mounted on an identical tower.

The maximum line-of-sight communication distance $$d$$ between two antennas of heights $$h_t$$ (transmitting) and $$h_r$$ (receiving) is given by the formula: $$d = \sqrt{2 R h_t} + \sqrt{2 R h_r}$$, where $$R$$ is the radius of the Earth.

Since the antennas are mounted on identical towers, their heights are equal: $$h_t = h_r = h$$. Substituting this into the distance formula gives: $$d = \sqrt{2 R h} + \sqrt{2 R h} = 2\sqrt{2 R h}$$.

We are given the following values:

Total separation distance, $$d = 45\text{ km} = 45 \times 10^3\text{ m}$$

Radius of the Earth, $$R = 6400\text{ km} = 6400 \times 10^3\text{ m} = 6.4 \times 10^6\text{ m}$$

Substituting these values into our simplified expression yields: $$45 \times 10^3 = 2\sqrt{2 \times (6.4 \times 10^6) \times h}$$.

Dividing both sides by 2 gives: $$22.5 \times 10^3 = \sqrt{12.8 \times 10^6 \times h}$$.

Squaring both sides to eliminate the square root results in: $$(22.5 \times 10^3)^2 = 12.8 \times 10^6 \times h$$, which simplifies to: $$506.25 \times 10^6 = 12.8 \times 10^6 \times h$$.

Canceling out $$10^6$$ from both sides leaves: $$506.25 = 12.8 \times h$$.

Solving for $$h$$ yields: $$h = \frac{506.25}{12.8} \approx 39.55\text{ m}$$.

Therefore, the correct answer is Option B: 39.55 m.