Let the mechanical energy be $E_{\text {mec }, t}$ when the child is at the top of the slide and $E_{\mathrm{mec}, b}$ when she is at the bottom. Then the conservation principle tells us
$$
E_{\mathrm{mec}, b}=E_{\mathrm{mec}, t} .
$$
8-3 READING A POTENTIAL ENERGY CURVE
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To show both kinds of mechanical energy, we have
$$
\begin{aligned}
K_b+U_b & =K_t+U_t, \\
\frac{1}{2} m v_b^2+m g y_b & =\frac{1}{2} m v_t^2+m g y_t .
\end{aligned}
$$
Dividing by $m$ and rearranging yield
$$
v_b^2=v_t^2+2 g\left(y_t-y_b\right)
$$
Putting $v_t=0$ and $y_t-y_b=h$ leads to
$$
\begin{aligned}
v_b & =\sqrt{2 g h}=\sqrt{(2)\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(8.5 \mathrm{~m})} \\
& =13 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
The huge advantage of using the conservation of energy instead of Newton's laws of motion is that we can jump from the initial state to the final state without considering all the intermediate motion. Here is an example. In Figure, a child of mass $m$ is released from rest at the top of a water slide, at height $h=8.5 \mathrm{~m}$ above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child's speed at the bottom of the slide.