Answer:
14.9 m/s
Explanation:
At the top of the hill, the skateboarder has both kinetic energy and potential energy, so its total mechanical energy is:
[tex]E=K_i+U_i = \frac{1}{2}mu^2 +mgh[/tex]
where
m = 82 kg is the mass of the skateboarder
u = 2.8 m/s is the initial velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h = 11 m is the initial height of the skateboarder
At the bottom of the hill, all the energy has been converted into kinetic energy, so:
[tex]E=K_f = \frac{1}{2}mv^2[/tex]
where
v is the final velocity of the skateboarder
Since the mechanical energy must be conserved, we can equate the two energies, and find the final velocity:
[tex]\frac{1}{2}mu^2 +mgh=\frac{1}{2}mv^2\\v=\sqrt{u^2+2gh}=\sqrt{2.8^2+2(9.8)(11)}=14.9 m/s[/tex]
