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Homework Help: Skier Jumps over a hedge!

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A skier swoops down a hill and over a ramp as in the attached figure. She starts from rest at a height of 16 m, leaves the 9.0 m ramp ata an angle of 45 degrees, and justt clears the hedge on her way down, making an angle of 30 degrees with the vertical as she does. Assuming that there is no friction, and that she is small compared to the dimensions of problem, solve for H, the height of the hedge in metres.

    See Diagram View attachment Diagram.doc

    Is it true that angle of decline and incline will have no effect in this problem?

    The answer is 2 m but I still don't known how to approach?

    2. Relevant equations

    KE = 0.5 * mass * (speed)^2

    PE = mass * g * height

    3. The attempt at a solution

    I don't know if this helps, but I think that roller coaster approach can be taken for this question.

    Which would mean that the mechanical energy will be conserves and the only form of force would be from gravity.

    We will also have Potential Energy (that will be max at top of curve and kenetic energy (that will be 0 at top and increase as skier comes down).

    KE = 0.5 * mass * (speed)^2

    PE = mass * g * height

    The skier leaves the ramp with an initial velocity of vi at an angle of 45 degrees to horizontal.
    Change in y=16-9=7m
    Gravitaional Potential=mgy
    Kenetic Energy=0.5mv^2

    At height of 9 m
    = 11.71 m/s

    Horizontal velocit is now 8.28 m/s
    Vertical velocity is 8.28 m/s too.

    I don't know how this will help in solving for the height of hedge. How would we use energy approach?

    I need help really soon
  2. jcsd
  3. Jan 28, 2007 #2
    Can some one please help! This is my second time posting the same question after four months, but everyone seems to be ignoring it. I haven't got a single reply! Thank You
  4. Jan 28, 2007 #3


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    I don't think everyone's ignoring you; personally, I was waiting for the picture to be approved before helping. You may have to wait till a mentor sees it!
  5. Jan 28, 2007 #4


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    This is all correct

    As for the second bit, I'm not sure whether you need to use energy methods; have you tried using the kinematic equations for this part?
  6. Jan 28, 2007 #5
    I did, but it did not help me!
    I got x=vi*t=8.28*1.689=14m
    Therefore horizontal distance =14m to the same level as ramp of exit height at 45 degrees.
    Now I had no idea how to ge the last bit where the angle of decline is 30 degrees. After this point I have no clue what's happening!

    Plus our teacher told us to use energy approach when marking the assignment.
  7. Jan 28, 2007 #6
    What are those angles? Are they the velocity vector's angle with horizontal? If that is the case, try using the equation of motion in 2-dimenion

    [tex] \textbf{r}(t) = (v_0\cos(\theta)t)\textbf{i} + (-\tfrac{1}{2}gt^2 + v_0\sin(\theta)t)\textbf{j}[/tex]

    where the origin of this coordinate system is at the point where "jump starts" at those 9m height.

    I see.. It is the angle of the velocity vectors you've been given. Find out at what time the velocity makes an angle of -30 deg with horizontal, and use that time to find what the height is by using the above vector equation.
    Last edited: Jan 28, 2007
  8. Jan 28, 2007 #7


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    To use energy, you need to equate the sum of the kinetic and potential energies when the skiier is at the hedge, minus the work done by gravity throughout the jump, to the total energy of the system.
  9. Jan 28, 2007 #8
    what does r, i and j refer to?

    how would i find out the time velocities and angle. I thought in parabola question the landing point has to be equal. Why is this below?
  10. Jan 28, 2007 #9
    How would I do that? How would I know the energies when skier is at the hedge. All I am given is that the angle =30 degrees. Nothing else. :confused:
  11. Jan 28, 2007 #10
    I assume you're not very familiar with basic vector calculus?
    Anyways, [itex]\textbf{r}[/itex] is the vector function describing the motion in 2-dimenion (direction). The net force on the skier is (in the xy-plan)

    [tex] \textbf{F} = \begin{pmatrix} 0 \\ -mg \end{pmatrix}[/tex]

    The vectors i, and j are unit vectors in the horizontal and vertical directions respectively, it's just another way of writting a vector. The above can be writtin as [itex]\textbf{F} = -mg\textbf{j}[/itex].
    From the acceleration vector, [itex] \textbf{a}=\textbf{F}/m[/itex], you can calculate what the vecloty and position vector of the motion would be, given the initial conditions.
    But I don't know how much you know about this.

    The "landing-point" can different, because it depends on where you start "on the parabola". Given the the equation [itex]y = ax^2 + y_0[/itex], if you start at [itex]x = 0[/itex], your (starting) height would be [itex] y = y_0 [/itex], and your "landing" point would be y = 0, which is different from your starting point.

    The velocity vector as a function of time in this case would be

    [tex] \textbf{v}(t) = \textbf{\dot{r}}(t) = \begin{pmatrix} v_0\cos\theta \\ -gt + v_0\sin\theta\end{pmatrix}[/tex]

    Where [itex]v_0[/itex] is the initial speed, that is the speed the skier have at the start of the jump, the angle [itex]\theta[/itex] is just the elevation for the initial velocity, the 45 deg.
    The velocity vector has to make an angle of -30 deg with horizontal at the heigh H you're seeking. How do you caculate the angle a vector makes with horizontal?
    Last edited: Jan 28, 2007
  12. Jan 28, 2007 #11
    This is a hopeless question. I think I will go and ask my teacher right before the exam how to do this. Thanks for all the help though! I am just too stupid for physics perhaps!
  13. Jan 28, 2007 #12


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    How precise is this answer? The closest I can get is about 2.8m, that's if I'm doing this right!
  14. Jan 28, 2007 #13
    We calculate it using trigonemetry, like sine, cosine law. and tan=opp/hyp and things like that.
  15. Jan 28, 2007 #14
    Don't you mean 2m ABOVE those 9 m? Otherwise I think I misinterpreted those angles. Aren't they the angles of the direction of the motion?
    If the that is the case, then the height can't be less than 9m, because of the following reason;
    The start angle (the 45 deg) which is the angle the velocity vector makes with horizontal, will decrease in time and reach 0 at the maximum height. Then it will start increasing in the negative direction. It will reach the same level of height (9m) only when the angle is 45 deg in the negative direction (-45 deg as a short notation). At an angle of -30 deg the skier would be at a greater height than those 9m it started from.
  16. Jan 28, 2007 #15
    This is a sir isaac newton contest question from Thomson Nelson Physics 12 textbook and the answer says exactly (2.0 m).
    I did this question 4 months ago and this is what I did and got 3/5:

    At height of 9 m
    = 11.71 m/s

    = vcostheta

    y=viyt+0.5ayt^2 (at this point teacher indicated that I should use y =vsin45t+0.5gt^2 energy approach)

    Because the skier comes down at 30 to vertical and 60deg to horizontal: y=xtan60

    Equating the equartions:

    solving for t I get 0.366

    then plugging this in to y equation

    = 11.7*0.366*0.5*9.8*0.66^2

    I get y to equation=2.8 m which is the wrong answer as indicated by the teacher. I did this question using solutions for newton contest with a similar question.
  17. Jan 28, 2007 #16
    Sorry!! I missed that part.
    Then I do get 2.01, which is the answer you're after.

    The method I told before is the same as what you are trying to do. I used vectors so that you could know how to use those 30 deg angles with vertical (or -60 deg with the horizontal).
    You know that the velocity vector has to make an angle of -60 deg with horizontal, you can calculate the angle of the vector using trig,

    [tex] \tan\phi = \frac{v_y}{v_x} = \frac{-gt+v_0\sin\theta}{v_0\cos\theta} [/tex]

    Those are the components of the velocity vector. Where [itex] \phi [/itex] is the angle with the vertical at time t. Note at t = 0, [itex] \phi = \theta [/itex]. Now at what time is [itex] \phi = -60^{\circ} [/itex]?

    Once you know this time you can calculate the height the skier has reached after this amount of time, since the equation of motion for the skier in the vertical direction is

    [tex] y(t) = -\tfrac{1}{2}gt^2 + v_yt + y_0 [/tex]
    Last edited: Jan 28, 2007
  18. Jan 28, 2007 #17
    Perhaps I made the hedge in the diagram too large. Too bad my scanner doesn't work other wise the pictre can be a lot more clearer. I kind of redid the diagram if that helps.
  19. Jan 28, 2007 #18

    This is a better diagram. Hope it helps

    Attached Files:

    Last edited: Jan 28, 2007
  20. Jan 28, 2007 #19
    Oh! Did you get the answer using what u told me before then?
  21. Jan 28, 2007 #20
    It's ok, it was my fault not reading you're #1 post more clear, since you did mention that it makes that angle with vertical.
    Read my last reply, and I'll try to figure out what you did wrong in your calculations.
  22. Jan 28, 2007 #21
    guess the diagram didn't work
  23. Jan 28, 2007 #22
  24. Jan 28, 2007 #23
    The direction of the acceleration is downwards. vsin45 is upwards, if you choose to a positive direction to be upwards, the acceleration should be negative, so a = -g.

    You can't say that tan60 = y/x, it's the change in y and x you need (velocity).
  25. Jan 28, 2007 #24
    No need to attach new diagrams, I understand the situation now.

    Read my reply at post #16.
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