The string of a kite is 100 m long and it makes an angle of 60° with the horizontal.

(a) $50 \sqrt{3} \mathrm{~m}$

Let $A B$ be the string of the kite and $A X$ be the horizontal line.

If $B C \perp A X$, then $A B=100 \mathrm{~m}$ and $\angle B A C=60^{\circ}$.

Let:

$B C=h \mathrm{~m}$

In the right $\triangle A C B$, we have:

$\frac{B C}{A B}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{h}{100}=\frac{\sqrt{3}}{2}$

$\Rightarrow h=\frac{100 \sqrt{3}}{2}=50 \sqrt{3} \mathrm{~m}$

Hence, the height of the kite is $50 \sqrt{3} \mathrm{~m}$.