91Ó°ÊÓ

The$3000\text{m}$ height mountain is located on the equator of the earth, the radius of earth is$6400\text{km}$.

The difference in speed is,

$\mathrm{Δ}v=v_1−v_2\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\mathrm{...}\left(\text{I}\right)$

Here, $\mathrm{Δ}v$is the difference in the speed, $v_1$is the speed of the climber in the top of the mountain, $v_2$is the speed of the surfer at nearby beach.

The speed of the climber in the top of the mountain is,

$v_1=\frac{2\mathrm{Ï€}(r+h)}{24\text{h}}$

Here, $r$is the radius of the earth, and $h$is the height of the mountain.

The speed of the surfer at a nearby beach is,

$v_2=\frac{2\mathrm{Ï€}r}{24\text{h}}$

Substitute $\frac{2\mathrm{π}(r+h)}{24\text{h}}$for $v_1$and $\frac{2\mathrm{π}r}{24\text{h}}$for $v_2$in the equation (I) to find $\mathrm{Δ}v$

$\mathrm{Δ}v=\frac{2\mathrm{π}(r+h)}{24\text{h}}−\frac{2\mathrm{π}r}{24\text{h}}$

$\frac{2\mathrm{π}\left(\left(6400\text{km}×\frac{1000\text{m}}{1\text{km}}\right)+3000\text{m}\right)}{\left(24\text{h}×\frac{60\min }{1\text{h}}×\frac{60\text{s}}{1\text{min}}\right)}−\frac{2\mathrm{π}\left(6400\text{km}×\frac{1000\text{m}}{1\text{km}}\right)}{\left(24\text{h}×\frac{60\text{min}}{1\text{h}}×\frac{60\text{s}}{1\text{min}}\right)}$

$\begin{array}{l}=465.63\text{m}/\text{s}−465.42\text{m}/\text{s}\\ =0.208\text{m}/\text{s}\end{array}$

Therefore, the speed of a climber on top of mountain is$0.208\text{m}/\text{s}$ faster than the speed of a surfer at nearby beach.