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(C) Given: $h_T = 50 \ m$,$h_R = 5 \ m$,$R = 6.4 	imes 10^6 \ m$. The formula for the maximum line-of-sight communication range $d$ is given by: $d =

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$33.3$

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(C) Given: $h_T = 50 \ m$,$h_R = 5 \ m$,$R = 6.4 	imes 10^6 \ m$. The formula for the maximum line-of-sight communication range $d$ is given by: $d = \sqrt{2h_T R} + \sqrt{2h_R R}$ Substituting the values: $d = \sqrt{2 	imes 50 	imes 6.4 	imes 10^6} + \sqrt{2 	imes 5 	imes 6.4 	imes 10^6}$ $d = \sqrt{640 	imes 10^6} + \sqrt{64 	imes 10^6}$ $d = 25.29 	imes 10^3 \ m + 8 	imes 10^3 \ m$ $d = 33.29 	imes 10^3 \ m \approx 33.3 \ km$.

$A$ transmitting antenna is at a height of $50 \ m$ and the receiving antenna is at a height of $5 \ m$. What is the maximum communication range in $km$? (Take $R = 6400 \ km$)

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