- #1

ec-physics

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- Homework Statement
- A skateboarder drops in from one end of a halfpipe, and travels to the top of the other end of the halfpipe. He needs two seconds to perform a twist.

Given the height of a halfpipe, and the mass of the skateboarder, what would be the skateboarder's maximum air time, and would he be able to perform his twist?

Halfpipe height h: 3.66m

Skateboarder mass m: 39.5 kg

- Relevant Equations
- PE=mgh

KE=1/2mv^2

vf = vi + at

I've calculated the potential energy at the top of the halfpipe, before the boarder drops in:

PE = 39.5 kg * 9.8 m/s^2 * 3.66 m = 1416 J

Since the boarder would have no potential energy and all kinetic energy at the bottom of the halfpipe,

KE = 1/2mv^2 = 1416 J

1/2 (39.5 kg) (v^2) = 1416 J

So his velocity at the bottom of the halfpipe is 8.46 m/s^2.

I'm getting stuck with translating that into the second half of the halfpipe journey, when he begins traveling up, catches air, and comes back down. Where do I go from here? I feel like I'm simplifying this into just kinetic and potential energy when there are more factors at play, like momentum.

PE = 39.5 kg * 9.8 m/s^2 * 3.66 m = 1416 J

Since the boarder would have no potential energy and all kinetic energy at the bottom of the halfpipe,

KE = 1/2mv^2 = 1416 J

1/2 (39.5 kg) (v^2) = 1416 J

So his velocity at the bottom of the halfpipe is 8.46 m/s^2.

I'm getting stuck with translating that into the second half of the halfpipe journey, when he begins traveling up, catches air, and comes back down. Where do I go from here? I feel like I'm simplifying this into just kinetic and potential energy when there are more factors at play, like momentum.