Skateboarder Needs Speed for 3m Quarter Pipe

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A 55kg skateboarder needs to achieve a specific speed at the bottom of a 3m quarter pipe to reach the upper edge. The necessary speed can be calculated using energy equations for potential and kinetic energy. By equating potential energy (mgh) to kinetic energy (mv²/2), the mass cancels out, simplifying the equation to v = sqrt(2gh). Substituting the values, with g as 9.81 m/s² and h as 3m, allows for the determination of the required speed. The discussion emphasizes the importance of understanding energy conservation in skateboarding dynamics.
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A 55kg skateboarder wants to make it to the upper edge of a quarter pipe with a radius 3.0 m. What speed do they need at the the bottom?
A.) 7.7 m/s
B.) 9.1 m/s
C.) 5.4 m/s

If you could help me out with the correct formula to use, that would help!
 
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Sorry new to the site... Revised the question on a later post
 


Hello Yomama

This can be done very quickly and effectivelly with the assistance of two energy equations:

Potential energy=mgh
Kinetinc Energy=(mv^2)/2

Equate the two in the following way:

mgh=(mv^2)/2 Note: m cancelles out

gh=(v^2)/2
2gh=v^2
v=sqrt(2gh) where: g=9.81, h=radius of pipe=3

Find v by substituting the value.

Hubert
 
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