Answer
After the moment it reached maximum height, it is falling; at $$t_{2}=t_{1}+2.5\mathrm{s}=5.5 \mathrm{s},$$
it will have fallen an amount given by Eq. $2-18$ :
$$
y_{\text {fence }}-y_{\max }=0-\frac{1}{2} g\left(t_{2}-t_{1}\right)^{2}
$$
Thus, the height of the fence is
$$
y_{\text {fence }}=y_{\max }-\frac{1}{2} g\left(t_{2}-t_{1}\right)^{2}=44.1 \mathrm{m}-\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(2.5 \mathrm{s})^{2}=13.48 \mathrm{m}
$$
Work Step by Step
After the moment it reached maximum height, it is falling; at $$t_{2}=t_{1}+2.5\mathrm{s}=5.5 \mathrm{s},$$
it will have fallen an amount given by Eq. $2-18$ :
$$
y_{\text {fence }}-y_{\max }=0-\frac{1}{2} g\left(t_{2}-t_{1}\right)^{2}
$$
Thus, the height of the fence is
$$
y_{\text {fence }}=y_{\max }-\frac{1}{2} g\left(t_{2}-t_{1}\right)^{2}=44.1 \mathrm{m}-\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(2.5 \mathrm{s})^{2}=13.48 \mathrm{m}
$$
