Let's use conservation of energy to find the velocity when the marble is launched:
[tex]\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ \frac{1}{2}mv1^2=\frac{1}{2}mv^2+mgh \\ mv1^2=mv^2+2mgh \\ v1^2=v2^2+2gh \\ solve_{\text{ }}for_{\text{ }}v2: \\ v2=\sqrt{v1^2-2gh} \\ v2=\sqrt{0.9978^2-2(9.8)(0)} \\ v2\approx0.9978m/s \end{gathered}[/tex]
Now, let's find the travel time as follows:
[tex]\begin{gathered} t=\frac{2v_0}{g} \\ t=\frac{2(0.9978)}{9.8} \\ t=0.204s \end{gathered}[/tex]
Now, we can find the vertical velocity as follows:
[tex]\begin{gathered} v_y=v_o+gt \\ so: \\ v_y=0.9978+(9.8)(0.204) \\ v_y=2.9934m/s \end{gathered}[/tex]
Answer:
2.9934 m/s
