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(B) The maximum range $d$ for line-of-sight communication is given by the formula: $d = \sqrt{2Rh_t} + \sqrt{2Rh_r}$,where $R$ is the radius of the Ea

Vedclass · Accepted Answer

$51.2$

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(B) The maximum range $d$ for line-of-sight communication is given by the formula: $d = \sqrt{2Rh_t} + \sqrt{2Rh_r}$,where $R$ is the radius of the Earth,$h_t$ is the height of the transmitting antenna,and $h_r$ is the height of the receiving antenna.Given: $d = 57.6 \ km = 57.6 	imes 10^3 \ m$,$h_r = 80 \ m$,$R = 6.4 	imes 10^6 \ m$.Substituting the values:$57.6 	imes 10^3 = \sqrt{2 	imes 6.4 	imes 10^6 	imes h_t} + \sqrt{2 	imes 6.4 	imes 10^6 	imes 80}$$57.6 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_t} + \sqrt{1024 	imes 10^6}$$57.6 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_t} + 32000$$57.6 	imes 10^3 - 32 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_t}$$25.6 	imes 10^3 = \sqrt{12.8 	imes 10^6 	imes h_t}$Squaring both sides:$(25.6 	imes 10^3)^2 = 12.8 	imes 10^6 	imes h_t$$655.36 	imes 10^6 = 12.8 	imes 10^6 	imes h_t$$h_t = \frac{655.36}{12.8} = 51.2 \ m$.

The maximum distance between the transmitting and receiving antennas for satisfactory communication in line of sight mode is $57.6 \ km$. If the height of the receiving antenna is $80 \ m$,the height of the transmitting antenna is (Radius of earth $= 6.4 \times 10^6 \ m$) (in $m$)

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