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Question

(B) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the form

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.Given: $d = 40 \ km = 40 	imes 10^3 \ m$,$h_R = 45 \ m$,$R = 6400 \ km = 6.4 	imes 10^6 \ m$.Substituting the values:$40 	imes 10^3 = \sqrt{2 	imes 6.4 	imes 10^6 	imes h_T} + \sqrt{2 	imes 6.4 	imes 10^6 	imes 45}$$40 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_T} + \sqrt{576 	imes 10^6}$$40 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_T} + 24 	imes 10^3$$40 	imes 10^3 - 24 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_T}$$16 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_T}$Squaring both sides:$(16 	imes 10^3)^2 = 12.8 	imes 10^6 	imes h_T$$256 	imes 10^6 = 12.8 	imes 10^6 	imes h_T$$h_T = \frac{256}{12.8} = 20 \ m$.

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