Answer:
About 7.67 m/s.
Explanation:
Mechanical energy is always conserved. Hence:
[tex]\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}[/tex]
Where U is potential energy and K is kinetic energy.
Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:
[tex]\displaystyle U_i = K_f[/tex]
Substitute and solve for final velocity:
[tex]\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}[/tex]
In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.
