Question

Calculation for the Final Vertical Velocity when marble becomes a horizontally launched projectile at the end of the roller coaster.

Answer 1

Let's use conservation of energy to find the velocity when the marble is launched:

`\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ (1)/(2)mv1^2=(1)/(2)mv^2+mgh \\ mv1^2=mv^2+2mgh \\ v1^2=v2^2+2gh \\ solve_{\text{ }}for_{\text{ }}v2: \\ v2=√(v1^2-2gh) \\ v2=√(0.9978^2-2(9.8)(0)) \\ v2\approx0.9978m/s \end{gathered}`

Now, let's find the travel time as follows:

`\begin{gathered} t=(2v_0)/(g) \\ t=(2(0.9978))/(9.8) \\ t=0.204s \end{gathered}`

Now, we can find the vertical velocity as follows:

`\begin{gathered} v_y=v_o+gt \\ so: \\ v_y=0.9978+(9.8)(0.204) \\ v_y=2.9934m/s \end{gathered}`

Answer:

2.9934 m/s