a) The child's speed halfway down the slide's vertical distance = 11.92 m/s
b) The child's speed three-fourths of the way down = 8.43 m/s
Explanation:The mass, m = 36 kg
The vertical height, h = 14.5 m
a) At halfway down the height, the kinetic energy and the potential energy are equal
[tex]\begin{gathered} \frac{1}{2}mv^2=mg(\frac{h}{2}) \\ \\ 0.5mv^2=m(g)(\frac{14.5}{2}) \\ \\ 0.5v^2=(9.8)(7.25) \\ \\ 0.5v^2=71.05 \\ \\ v^2=\frac{71.05}{0.5} \\ \\ v^2=142.1 \\ \\ v=\sqrt{142.1} \\ \\ v=11.92\text{ m/s} \end{gathered}[/tex]b) To get the child's speed three-fourths of the way down
[tex]\begin{gathered} mg(h-\frac{3h}{4})=\frac{1}{2}mv^2 \\ \\ mg(\frac{h}{4})=m\frac{v^2}{2} \\ \\ v^2=g\frac{h}{2} \\ \\ v^2=\frac{9.8(14.5)}{2} \\ \\ v^2=71.05 \\ \\ v=\sqrt{71.05} \\ \\ v=8.43\text{ m/s} \end{gathered}[/tex]The child's speed three-fourths of the way down = 8.43 m/s
