A TV transmission tower antenna is at a height of 20 m. Suppose that the receiving antenna is at (i) ground level (ii) a height of 5 m. The increase in antenna range in case (ii) relative to case (i) is $$n$$%. The value of $$n$$, to the nearest integer, is ___.

The range of a TV transmission tower of height $$h_T$$ with a receiving antenna at height $$h_R$$ is given by $$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$, where $$R$$ is the radius of the Earth.

In case (i), the receiving antenna is at ground level ($$h_R = 0$$), so the range is $$d_1 = \sqrt{2R \times 20} = \sqrt{40R}$$.

In case (ii), the receiving antenna is at height 5 m, so the range is $$d_2 = \sqrt{2R \times 20} + \sqrt{2R \times 5} = \sqrt{40R} + \sqrt{10R}$$.

The percentage increase in range is $$n = \frac{d_2 - d_1}{d_1} \times 100 = \frac{\sqrt{10R}}{\sqrt{40R}} \times 100 = \frac{1}{2} \times 100 = 50\%$$.

The value of $$n$$ is $$50$$.