Jumping straight up while skateboarding downhill

[Renoir]

Hi,

I understand the mechanics of jumping straight up and landing on the same spot when inside of a moving bus traveling at a constant velocity but what happens when I jump straight up when skateboarding downhill? Do I land on the same spot on the skateboard, do I land further back relative to the skateboard etc? What forces are involved and what is the explanation for what happens?

Thanks very much in advance for your replies!

Renoir

 

[A.T.]

when I jump straight up when skateboarding downhill?

Accelerate vertically against gravity, while keeping the current horizontal speed? This might be tricky to achieve, and might accelerate the skateboard. You have to describe this more precisely.

Do I land on the same spot on the skateboard, do I land further back relative to the skateboard etc?

Is the skateboard's horizontal velocity component constant? What about yours?

 

[Renoir]

Accelerate vertically against gravity, while keeping the current horizontal speed? This might be tricky to achieve, and might accelerate the skateboard. You have to describe this more precisely.

Is the skateboard's horizontal velocity component constant? What about yours?

Yes vertically against gravity while keeping downhill speed constant (it being tricky to jump vertically without accelerating the skateboard we can ignore for this question). Skateboard & I are going downhill at constant speed when I jump up.

 

[A.T.]

Yes vertically against gravity while keeping downhill speed constant (it being tricky to jump vertically without accelerating the skateboard we can ignore for this question). Skateboard & I are going downhill at constant speed when I jump up.

Then replace the skateboard with an escalator or sloped moving walkway. It should make it clearer what happens.

 

[jkn]

Hi,

I understand the mechanics of jumping straight up and landing on the same spot when inside of a moving bus traveling at a constant velocity but what happens when I jump straight up when skateboarding downhill? Do I land on the same spot on the skateboard, do I land further back relative to the skateboard etc? What forces are involved and what is the explanation for what happens?

Thanks very much in advance for your replies!

Renoir

If you want to maintain your balance, you have to jump perpendicular to the road, not straight up. Small problem remains: While you are flying you accelerate only down. Your skateboard might keep accelerating forward, because of gravity and downhill.

 

[Renoir]

If you want to maintain your balance, you have to jump perpendicular to the road, not straight up. Small problem remains: While you are flying you accelerate only down. Your skateboard might keep accelerating forward, because of gravity and downhill.

Thanks for the reply!
So if I jump straight up then I will land further back relativ to skateboard but further forward relativ to ground? If so is that because the jump directly up is somewhat (depending on slope of hill) in opposite direction to direction of skateboards travel where as jumping perpendicular to skateboard (so diagonally in air) would effectively not change the forward momentum of my body similar to jumping in a bus driving on flat road?

 

[A.T.]

So if I jump straight up then I will land further back relativ to skateboard but further forward relativ to ground?

Not given the unrealistic assumptions you make in post #3. But in reality yes, it's like jumping backwards from the skateboard, because the vertical jump direction, projected onto the slope points uphill.

 

[sophiecentaur]

Your skateboard might keep accelerating forward, because of gravity and downhill.

The component of your own acceleration downhill will be the same as for the board. The difference could be that the rolling friction of the board could reduce its acceleration to a value below yours, so the landing point could be slightly forward on the board. But in 1s, the difference is unlikely to constitute a significant error on a smooth road with good bearings. With no friction involved, the landing would be in the same spot but the drag forces on the board would affect the board's speed mor when you are not on it than when you are on it.
The jump would need to be normal to the road surface so there would, ideally, be no change in velocity parallel to the road for you or the board.

 

[zanick]

The component of your own acceleration downhill will be the same as for the board. The difference could be that the rolling friction of the board could reduce its acceleration to a value below yours, so the landing point could be slightly forward on the board. But in 1s, the difference is unlikely to constitute a significant error on a smooth road with good bearings. With no friction involved, the landing would be in the same spot but the drag forces on the board would affect the board's speed mor when you are not on it than when you are on it.
The jump would need to be normal to the road surface so there would, ideally, be no change in velocity parallel to the road for you or the board.

drag when you are not coupled to the skateboard will be different for you and the skateboard. if the drag is not equal, one will slow faster than the other, so the spot you land on willl not be the same at which you departed from. ;) (ignoring the forward or rearward forces imparted on the board upon jumping)

 

[sophiecentaur]

I thought that was what I said. A good diagram of situations like this can help avoid misunderstandings. IMO, diagrams should accompany many questions

 

[Cutter Ketch]

I think AT had a very good idea. If you start throwing in what the board does without thinking about what you do separate from the board this will get very confused.

So suppose you are on a sloped conveyor belt moving at skate board speed. Your velocity vector is down the hill. The gravity vector has a component down the hill so to keep from slipping you are relying on a friction vector pointing up the hill. You jump straight up against gravity. The normal force is perpendicular to the belt, so to accelerate straight up you must also increase the friction force up the hill. That will be important when we switch to the skateboard. Once off of the belt your horizontal speed remains constant. Here is the part I think drove the question: if the belt is moving a constant speed, the horizontal component of its speed is also constant. When you jump straight up you still stay above the same point on the belt. The fact that it is sloped just means that it falls away from you and so you have to fall longer and travel further to get back to it.

In addition you experience air resistance, but for our short leap at low speed let's say that is negligible.

Now let's add the skateboard. To jump straight up you need a component of force parallel to the slope. The skateboard won't provide it. If you try to jump straight up you will kick the skateboard out in front of you. To within the small amount of friction and inertia the board can provide you pretty much have to jump perpendicular to the slope. This means that you increase your horizontal velocity. If the board stayed at a constant velocity you would leap in front of it. However the board doesn't stay at a constant velocity. It continues to accelerate down the hill. Once you leave the board your horizontal component is larger but constant. The board's horizontal component is smaller but growing and there is the possibility that you will come back together if you jump just the right amount.

Does that mean there is one and only one jump that returns to the board? No! Remember if you don't jump perfectly perpendicular to the slope you kick the board forward or back? You can angle your jump to change the speed of the board to make a whole range of jumps work out.

 

[sophiecentaur]

So suppose you are on a sloped conveyor belt moving at skate board speed.

Why introduce this complication? It's not part of the OP.
In the simplest situation, the skateboard will be accelerating and so will you. Your two accelerations vectors, downhill will be the same. If you jump normal to the slope your accelerate, velocities and distance traveled will be the same of you and the board. Your body describes a parabola and board travels in a straight line. They will intersect when you return to the ground.
Nearer 'real life', the board will not be accelerating as much as you, because of rolling resistance. Just how much, will depend on the situation and it would be possible to choose a takeoff angle to ensure landing back on the board but that would involve your calculator whilst you are on the move. The board could actually be at its terminal velocity, once you have jumped off - or even slow up a bit / stop. It's much to complicated to predict without a lot more data.

 

[A.T.]

Why introduce this complication? It's not part of the OP.

Replacing the skateboard with a constant speed conveyor belt is a simplification, not a complication. I proposed it based on the simplifying assumptions sated in post #3. But I'm not sure if I understood post #3 correctly: Constant speed throughout the jump or just before the jump?

If you jump normal to the slope...

That is explicitly not the question (see post #3).

 

[sophiecentaur]

That is explicitly not the question (see post #3).

Maybe not but, if you want to achieve the aim of landing on the board, you have either to specify the whole system in much more detail or make assumptions that the setup is ideal. In that case, you have to jump Normal to the slope and then it works. Without more numbers, there is no solution.
I have already made the point that the problem is not specified well enough discuss without confusion.

 

[sophiecentaur]

If you do not jump along the Normal, you need to know the mass of the skateboard and the person, to find the relative velocities. etc. etc.

 

[jkn]

The component of your own acceleration downhill will be the same as for the board. The difference could be that the rolling friction of the board could reduce its acceleration to a value below yours, so the landing point could be slightly forward on the board. But in 1s, the difference is unlikely to constitute a significant error on a smooth road with good bearings. With no friction involved, the landing would be in the same spot but the drag forces on the board would affect the board's speed mor when you are not on it than when you are on it.
The jump would need to be normal to the road surface so there would, ideally, be no change in velocity parallel to the road for you or the board.

When skater is in the air, during the jump, component of his acceleration forward is zero (ignoring air resistance and wind). Skateboard might accelerate forward because of downhill and gravity. It might deaccelerate because of rolling resistance. If jumper is lucky those forces cancel each other and he lands on the board.

When jumping on level ground, rolling resistance will slow skateboard down. Skater has learned to compensate. In downhill necessary compensation is smaller or to other direction.

 

[sophiecentaur]

When skater is in the air, during the jump, component of his acceleration forward is zero

The word "forward" needs to be defined. It could mean 'horizontal' or ' along the plane of the slope'. It's motion along the plane of the slope that is of interest here.
Gravity acts the same on the skater, whether he is on the ground or in the air. His weight force is vertically down. This can be resolved into two forces - one, normal to the slope and another one along the plane of the slope. The force down the slope is not zero so his motion along the plane of the slope is actually accelerating all the time - just the same as the board with no friction. (This must be sorted out before introducing any friction or a non-normal takeoff angle.)

 

[sophiecentaur]

The jumping skater actually follows a Parabolic trajectory for non-vertical takeoff.

 

[jkn]

The word "forward" needs to be defined. It could mean 'horizontal' or ' along the plane of the slope'. It's motion along the plane of the slope that is of interest here.
Gravity acts the same on the skater, whether he is on the ground or in the air. His weight force is vertically down. This can be resolved into two forces - one, normal to the slope and another one along the plane of the slope. The force down the slope is not zero so his motion along the plane of the slope is actually accelerating all the time - just the same as the board with no friction. (This must be sorted out before introducing any friction or a non-normal takeoff angle.)

Problem is simpler if we use forward at straight angle with up. Of course acceleration can be divided as you write. With it you arrived incorrect conclusion earlier: "In the simplest situation, the skateboard will be accelerating and so will you. Your two accelerations vectors, downhill will be the same."

So I insist using simpler approach to get to correct answer. Difference between jump from moving skateboard on level ground and on downhill is important, because skater probably practices jumps on level ground first.

Rolling resistance is same assuming same speed. In both cases jumper maintains constant speed forward and accelerates only downwards while flying. In downhill there is an extra force forward on skateboard. So jumper how has practiced on level ground might land too far back on the skateboard.

Yes trajectory is parabolic as seen by standing observer, but that detail does not help to answer original question.

 

[jbriggs444]

Problem is simpler if we use forward at straight angle with up. Of course acceleration can be divided as you write. With it you arrived incorrect conclusion earlier: "In the simplest situation, the skateboard will be accelerating and so will you. Your two accelerations vectors, downhill will be the same."

Which coordinate system is simpler depends on which specific problem and which aspect of that problem one is addressing. For a question of whether the jumper's trajectory eventually intersects with a frictionless skateboard, I agree with @sophiecentaur -- placing the x-axis parallel to the slope is simpler. One can read off the result without any calculation at all. The accelerations "downhill" (parallel to the surface of the slope) are indeed identical and it all comes down to launch angle.

A jump normal to the slope works. A vertical jump fails.

 

[jkn]

Which coordinate system is simpler depends on which specific problem and which aspect of that problem one is addressing. For a question of whether the jumper's trajectory eventually intersects with a frictionless skateboard, I agree with @sophiecentaur -- placing the x-axis parallel to the slope is simpler. One can read off the result without any calculation at all. The accelerations "downhill" (parallel to the surface of the slope) are indeed identical and it all comes down to launch angle.

A jump normal to the slope works. A vertical jump fails.

Jump has to be very close to normal. Giving board little speed forward or backward might be possible.

In this case vertical-horizontal coordinates are simpler.

Which of these claims you disagree with:

1: Object in free fall will ONLY accelerate downwards. (Ignoring air resistance and Coriolis force which are very small in this case.)

2: Friction less skateboard in downhill will accelerate in horizontal direction. Of course also downwards.

 

[jbriggs444]

Jump has to be very close to normal. Giving board little speed forward or backward might be possible.

In this case vertical-horizontal coordinates are simpler.

Which of these claims you disagree with:

1: Object in free fall will ONLY accelerate downwards. (Ignoring air resistance and Coriolis force which are very small in this case.)

2: Friction less skateboard in downhill will accelerate in horizontal direction. Of course also downwards.

Both claims are correct, of course. But which of these claims do you disagree with?

1: The downward accelerations of skateboard and jumper are different.
2: The parallel-to-the-slope accelerations of [frictionless] skateboard and jumper are identical.

 

[sophiecentaur]

Problem is simpler if we use forward at straight angle with up.

It may be simpler for you, because you may know what that means. If you use terms like normal, vertical, horizontal etc. then everyone will know what you mean.

Yes trajectory is parabolic as seen by standing observer,

Where is more convenient than the Earth's frame of reference? You are discussing this from the standpoint of a skater and that includes many more variables which you would need to specify and to include in any calculation - if you want a discussion of the Physics.
I have given the only solution that's available and it applies to the only possible scenario that your initial (limited) conditions describes. My solution is correct for that. If you want more details then you must specify more.

A jump normal to the slope works. A vertical jump fails.

I would go further than that. Any jump that's not normal to the slope cannot be relied on to work. It will cause the board to change velocity - faster or slower, depending on the angle. My solution would work (frictionless situation) whatever the takeoff velocity happened to be - he will meet the board after a leap of 100m/s - as long as the slope is long enough.

 

[jkn]

Both claims are correct, of course. But which of these claims do you disagree with?

1: The downward accelerations of skateboard and jumper are different.
2: The parallel-to-the-slope accelerations of [frictionless] skateboard and jumper are identical.

Of course I have to accept both. I found an error from my logic. No, I still don't believe that free falling objects accelerate sideways. Jump must be into direction of normal. So it has horizontal component. In one example case it matched with horizontal acceleration of the skate board. So that jumper lands on same spot on skateboard he jumped from. (I should calculate properly, but not now.) I still think there is a difference with jumps with or without downhill. I should calculate before posting and I don't have time now.

 

[sophiecentaur]

No, I still don't believe that free falling objects accelerate sideways.

That sounds like reasonable statement except that 'sideways' down the slope has a vertical component. If your non-belief was valid then there would be nothing to make the board go downhill, either.
I made an earlier comment about using the right terms and "sideways" is not a defined term. You have to specify whether sideways means horizontal or parallel to the plane of the slope. There is a significant difference, which you would see more obviously if you worked with a very steep slope. - nearly vertical, for instance.

 

[jbriggs444]

I should calculate before posting and I don't have time now.

If one uses the parallel-to-the-slope coordinate system is then no calculation is required. There is only one force acting in a direction parallel to the slope. It acts on jumper and skate board identically. It follows that the jumping in the normal direction results in landing on the board.

That's simple.

 

[sophiecentaur]

I still think there is a difference with jumps with or without downhill. I should calculate before posting and I don't have time now.

If you calculate for a slope, starting with a formula, you only need to put the value of the slope to zero to find the horizontal condition. I think you must be a skater and that you are letting your perceptions of your experience rule the basic physics. Your real situations involve many more variables than what we have been discussing. I put it to you that you would actually have no idea of precisely the angle you are jumping at. It's just a feeling you have. That's essential for good skating, of course, but clouds the Physics discussion. To prove all this stuff for yourself you would have to make a simulation, with a spring and a barrel, with adjustable elevation, firing a ball. The ball would behave pretty much as we are predicting. (Smooth slope and good bearings etc. etc.) You could probably win several bets with your skater colleagues. It would be a similar situation to Galileo's gravity experiment with the Tower of Pisa, where no one could believe him.