The maximum distance between the transmitting | vedclass

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Question

(B) The maximum line-of-sight distance $d$ between a transmitting tower of height $h_T$ and a receiving tower of height $h_R$ is given by $d = \sqrt{2

Vedclass · Accepted Answer

$70.3 \,m, 95.7 \,m$

Vedclass · Answer

(B) The maximum line-of-sight distance $d$ between a transmitting tower of height $h_T$ and a receiving tower of height $h_R$ is given by $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$, where $R$ is the radius of the Earth.Given: $d = 65 \,km = 65000 \,m$, $R = 6400 \,km = 6.4 	imes 10^6 \,m$, and $\frac{h_T}{h_R} = \frac{36}{49}$.From the ratio, let $h_T = 36k$ and $h_R = 49k$.Substituting these into the distance formula:$65000 = \sqrt{2 	imes 6.4 	imes 10^6 	imes 36k} + \sqrt{2 	imes 6.4 	imes 10^6 	imes 49k}$$65000 = \sqrt{2 	imes 6.4 	imes 10^6} 	imes (6\sqrt{k} + 7\sqrt{k})$$65000 = \sqrt{12.8 	imes 10^6} 	imes 13\sqrt{k}$$65000 = 3577.7 	imes 13\sqrt{k}$$65000 = 46510.1 	imes \sqrt{k}$$\sqrt{k} = \frac{65000}{46510.1} \approx 1.3975$$k \approx 1.953$$h_T = 36 	imes 1.953 \approx 70.3 \,m$$h_R = 49 	imes 1.953 \approx 95.7 \,m$.

The maximum distance between the transmitting and receiving $TV$ towers is $65 \,km$. If the ratio of the heights of the $TV$ transmitting tower to receiving tower is $36: 49$, the heights of the transmitting and receiving towers respectively are (radius of earth $= 6400 \,km$)

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