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B Jumping straight up while skateboarding downhill

  1. Dec 5, 2016 #1
    Hi,

    I understand the mechanics of jumping straight up and landing on the same spot when inside of a moving bus travelling 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
     
  2. jcsd
  3. Dec 5, 2016 #2

    A.T.

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    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?
     
  4. Dec 5, 2016 #3
    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.
     
  5. Dec 5, 2016 #4

    A.T.

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    Then replace the skateboard with an escalator or sloped moving walkway. It should make it clearer what happens.
     
  6. Dec 5, 2016 #5

    jkn

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    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.
     
  7. Dec 5, 2016 #6
    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?
     
  8. Dec 5, 2016 #7

    A.T.

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    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.
     
  9. Dec 5, 2016 #8

    sophiecentaur

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    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.
     
  10. Dec 10, 2016 #9
    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)
     
  11. Dec 11, 2016 #10

    sophiecentaur

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    I thought that was what I said. A good diagram of situations like this can help avoid misunderstandings. IMO, diagrams should accompany many questions
     
  12. Dec 11, 2016 #11
    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.
     
  13. Dec 11, 2016 #12

    sophiecentaur

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    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 travelled 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.
     
  14. Dec 11, 2016 #13

    A.T.

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    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?

    That is explicitly not the question (see post #3).
     
  15. Dec 11, 2016 #14

    sophiecentaur

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    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.
     
  16. Dec 11, 2016 #15

    sophiecentaur

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    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.
     
  17. Dec 12, 2016 #16

    jkn

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    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.
     
  18. Dec 12, 2016 #17

    sophiecentaur

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    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.)
     
  19. Dec 12, 2016 #18

    sophiecentaur

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    The jumping skater actually follows a Parabolic trajectory for non-vertical takeoff.
     
  20. Dec 12, 2016 #19

    jkn

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    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.
     
  21. Dec 12, 2016 #20

    jbriggs444

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    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.
     
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