# Airtime for halfpipe skateboarder

• ec-physics
In summary: A= (v^2/r)^2, where r is the radius of the halfpipeA= (8.46 m/s^2*3.66 m)*(9.8 m/s^2/r)^2 = 5.89 m/s^2
ec-physics
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.

You should keep symbolic quantities symbolic as long as possible. This makes it easier to do the algebra. If you had done this you could have noticed that the skater's mass is irrelevant. You multiplied and divided by 39.5 for no reason.

As the skater comes up the other side of the half pipe, his kinetic energy is converted to potential. How much kinetic energy should he have left when he returns to the original starting height?

Edit: It seems that the problem statement is missing information. I can see a way to solve if we make an assumption about the means by which the skater gains kinetic energy (he rises from a crouch at the very bottom of the half-pipe), an assumption about the half-pipe profile (semi-circle, vertical ends and radius 3.66 meters), an assumption about how many cycles he has to gain speed (one pass), an assumption about his starting velocity (starting from rest) and an assumption about how deeply he is able to crouch (say 0.5 meters).

Last edited:
jbriggs444 said:
You should keep symbolic quantities symbolic as long as possible. This makes it easier to do the algebra. If you had done this you could have noticed that the skater's mass is irrelevant. You multiplied and divided by 39.5 for no reason.

As the skater comes up the other side of the half pipe, his kinetic energy is converted to potential. How much kinetic energy should he have left when he returns to the original starting height?

Since he's back to the same height, at TE=PE+KE, he should have no kinetic energy and all potential, right?
So in theory he wouldn't ever get above the 3.66m halfpipe?

This is why I'm pretty sure I'm missing something, or pursuing the wrong avenue.

ec-physics said:
Since he's back to the same height, at TE=PE+KE, he should have no kinetic energy and all potential, right?

So in theory he wouldn't ever get above the 3.66m halfpipe?
Right.
This is why I'm pretty sure I'm missing something, or pursuing the wrong avenue.
See my edited response above for an avenue by which a solution might be pursued.

jbriggs444 said:
Right.

See my edited response above for an avenue by which a solution might be pursued.
So if he does change his height, that would change the radius from the center of the halfpipe (assuming it's a semi-circle), which would change... angular velocity? I've looked at this link briefly: https://www.real-world-physics-problems.com/physics-of-skateboarding.html
Think it could point me down the right road to finding out air time?

ec-physics said:
So if he does change his height, that would change the radius from the center of the halfpipe (assuming it's a semi-circle), which would change... angular velocity? I've looked at this link briefly: https://www.real-world-physics-problems.com/physics-of-skateboarding.html
Think it could point me down the right road to finding out air time?
Right. That is one way to attack the question of how much kinetic energy is gained. Rising from the crouch changes his "moment of inertia". Angular momentum is conserved, so angular velocity must change inversely.

That gives you a starting point with which to compute the resulting velocity and energy.

A different approach which should yield the same final result would consider the work done by the skater's legs lifting himself against the centrifugal force field.

jbriggs444 said:
Right. That is one way to attack the question of how much kinetic energy is gained. Rising from the crouch changes his "moment of inertia". Angular momentum is conserved, so angular velocity must change inversely.

That gives you a starting point with which to compute the resulting velocity and energy.

A different approach which should yield the same final result would consider the work done by the skater's legs lifting himself against the centrifugal force field.
Question: how would I go about considering the amount of work done by the skater's legs? Would that be just an estimate based on how much a person can push with their legs?

ec-physics said:
Question: how would I go about considering the amount of work done by the skater's legs? Would that be just an estimate based on how much a person can push with their legs?
You know the velocity at the bottom of the curve.
That tells you the centripetal acceleration at the bottom of the curve (and along trajectories with narrower radii)
If the person has the strength to rise out of the crouch, that tells you how much force he needs to push with. A difficulty is that the tangential velocity will change due to the skater's efforts. So the centrifugal force field does not have a simple potential function. You probably do not want to go this way.

The angular momentum approach is much simpler. You can get the resulting velocity change almost immediately.

Simple calculation: In 2 seconds an object released from rest drops distance
$$d=\frac{1}{2}gt^2=\frac{1}{2}\times 9.8~ \rm{(m/s^2)}\times 2^2~\rm{(s^2)}=19.6 \rm~{m}$$
This means that he needs to rise at least up to that height above the bottom of the half pipe. I am not a skateboarder but I think 2 seconds of air time is unrealistic.

kuruman said:
Simple calculation: In 2 seconds an object released from rest drops distance
$$d=\frac{1}{2}gt^2=\frac{1}{2}\times 9.8~ \rm{(m/s^2)}\times 2^2~\rm{(s^2)}=19.6 \rm~{m}$$
This means that he needs to rise at least up to that height above the bottom of the half pipe. I am not a skateboarder but I think 2 seconds of air time is unrealistic.
You can cut that vertical height by a factor of four if you realize that two seconds hang time is one second up and one second back down. It's still pretty high, but not outlandishly so.

## 1. How is airtime measured for a halfpipe skateboarder?

Airtime for a halfpipe skateboarder is typically measured by the amount of time the skateboarder spends in the air before landing back on the ramp. This can be measured using a stopwatch or timing device.

## 2. What factors affect the amount of airtime a halfpipe skateboarder can achieve?

The amount of airtime a halfpipe skateboarder can achieve is affected by a variety of factors, including the speed and height at which they enter the ramp, the angle of the ramp, and the strength and technique of the skateboarder.

## 3. How do halfpipe skateboarders increase their airtime?

Halfpipe skateboarders can increase their airtime by building up more speed before entering the ramp, using their body and legs to generate more lift, and mastering advanced tricks and maneuvers that allow them to stay in the air for longer periods of time.

## 4. What is the average airtime for a halfpipe skateboarder?

The average airtime for a halfpipe skateboarder can vary depending on the skill level of the skateboarder and the size and shape of the ramp. However, experienced skateboarders can typically achieve airtime of 3-5 seconds, while professional skateboarders can reach airtime of 7-8 seconds or more.

## 5. Is airtime important for a halfpipe skateboarder?

Airtime is a crucial aspect of halfpipe skateboarding as it allows skateboarders to perform tricks and maneuvers that require them to be in the air. It also adds an element of excitement and difficulty to the sport, making it more enjoyable for both the skateboarder and the audience.

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