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(B) For line-of-sight communication between two antennas of height $h$,the maximum distance $D$ is given by the formula $D = \sqrt{2Rh_t} + \sqrt{2Rh_

Vedclass · Accepted Answer

$39.55$

Vedclass · Answer

(B) For line-of-sight communication between two antennas of height $h$,the maximum distance $D$ is given by the formula $D = \sqrt{2Rh_t} + \sqrt{2Rh_r}$.Since the towers are identical,$h_t = h_r = h$,so $D = 2\sqrt{2Rh}$.Given $D = 45 \, km$ and $R = 6400 \, km$.Squaring both sides,we get $D^2 = 4(2Rh) = 8Rh$.Therefore,$h = \frac{D^2}{8R}$.Substituting the values: $h = \frac{45^2}{8 	imes 6400} \, km = \frac{2025}{51200} \, km$.$h \approx 0.03955 \, km = 39.55 \, m$.

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

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