In $\triangle \mathrm{ABC}, \mathrm{AC}$ represents ladder
$\mathrm{AB} \rightarrow$ vertical wall
Let $\mathrm{AB}=x, \mathrm{BC}=y$
$\therefore \quad \angle \mathrm{ABC}=90^{\circ}$
By Pythagoras theorem,
$\begin{aligned}
& \mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2 \\
& \Rightarrow x^2+y^2=17^2 \\
& \Rightarrow x^2=289-y^2 ... (i)\\
& \Rightarrow x^2=289-64 \\
& \Rightarrow x^2=225 \\
& \Rightarrow x=15 \mathrm{~m}
\end{aligned}$
Consider equation (i),
$x^2=289-y^2$
Differentiating w.r.t. t, we get
$\begin{aligned}
& 2 x \frac{\mathrm{d} x}{\mathrm{dt}}=-2 y \frac{\mathrm{d} y}{\mathrm{dt}} \\
& \Rightarrow 15 \frac{\mathrm{d} x}{\mathrm{dt}}=-8(1) \\
& \Rightarrow \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{-8}{15} \mathrm{~m} / \mathrm{s}
\end{aligned}$
Negative sign shows that the ladder is moving down. i.e., vertical length is decreasing
$\therefore \quad$ Upper end is coming down at the rate of $\frac{8}{15} \mathrm{~m} / \mathrm{s}$.