Answer:
The velocity of the skateboarder at point B = 6.02 m/s
Explanation:
For the motion down the hill
[tex]\begin{gathered} The\text{ intial velocity, u = 1.7 m/s} \\ \\ g=9.8\text{ m/s}^2 \\ \\ The\text{ height, H}_A\text{ = 2.7 m} \end{gathered}[/tex][tex]\begin{gathered} v^2=u^2+2gH_A \\ \\ v^2=1.7^2+2(9.8)(2.7) \\ \\ v^2=55.81 \\ \\ v=\sqrt{55.81} \\ \\ v=7.47\text{ m/s} \end{gathered}[/tex]
For the motion up the hill
The final velocity of the last motion will now be the initial velocity
[tex]\begin{gathered} u=7.47 \\ \\ H_B=1.0m \\ \\ g=-9.8\text{ m/s}^2(upward\text{ motion\rparen} \\ \\ \end{gathered}[/tex]
The velocity at B is calculated as:
[tex]\begin{gathered} v_B^2=u^2+2gH_B \\ \\ v_B^2=7.47^2+2(-9.8)(1) \\ \\ v_B^2=36.2 \\ \\ v_B=\sqrt{36.2} \\ \\ v_B=6.02\text{ m/s} \end{gathered}[/tex]
The velocity of the skateboarder at point B = 6.02 m/s
