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(B) Given: Height of transmitting antenna $h_T = 50\, m$ and height of receiving antenna $h_R = 32\, m$.Radius of the Earth $R \approx 6.4 	imes 10^6

Vedclass · Accepted Answer

$45.5$

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(B) Given: Height of transmitting antenna $h_T = 50\, m$ and height of receiving antenna $h_R = 32\, m$.Radius of the Earth $R \approx 6.4 	imes 10^6\, m$.The maximum line-of-sight distance $d_M$ is given by the formula: $d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.Substituting the values:$d_M = \sqrt{2 	imes 6.4 	imes 10^6 	imes 50} + \sqrt{2 	imes 6.4 	imes 10^6 	imes 32}$.$d_M = \sqrt{640 	imes 10^6} + \sqrt{409.6 	imes 10^6}$.$d_M = (25.298 	imes 10^3) + (20.238 	imes 10^3) = 45.536 	imes 10^3\, m$.Converting to kilometers: $d_M \approx 45.5\, km$.

$A$ transmitting antenna at the top of a tower has height $32\, m$ and the height of the receiving antenna is $50\, m$. What is the maximum distance between them for satisfactory communication in line of sight $(LOS)$ mode? ........$km$

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