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Easy problem 2 answers

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    An astronaut on a strange planet finds that she can jump a max horizontal distance of 15m if her initial speed is 3m/s. What is the free-fall acceleration on the planet?


    2. Relevant equations



    3. The attempt at a solution
    [tex]2a(x_{f}-x_{i})=v^{2}_{f}-v^{2}_{i}[/tex]
    [tex]\frac{2a(15)}{30}=9\\a=-.3m/s^{2}[/tex]
    OR
    [tex]\Delta x=\frac{v^{2}_{i} sin(2\theta)}{a}\\ \therefore a=\frac{v^{2}}{\Delta x}\\ a=-.6m/s^{2}[/tex]


    ...Confused :s
     
  2. jcsd
  3. Feb 11, 2013 #2

    cepheid

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    The first method is wrong because it's just considering motion in one dimension. I.e. if your horizontal acceleration were a, and you moved a distance Δx, starting at speed vi, your final speed would be vf. That's what that equation says.

    For projectile motion, your horizontal acceleration is 0, and your vertical acceleration is g, which is what you are trying to solve for. So the second method, which gives the horizontal range of a projectile given acceleration and launch speed and angle, is what you want. Clearly the launch angle is not straight up, otherwise you wouldn't move horizontally at all. So setting sin(2θ) = 1 was the wrong thing to do. Hint: the range is maximized when θ = 45°
     
  4. Feb 11, 2013 #3
    The equation on my sheet says that's the equation for it. If I don't use sin2x then I get -.424m/s^2
     
  5. Feb 11, 2013 #4

    cepheid

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    Yeah I made a mistake. It's 2θ, not θ, so when θ = 45°, 2θ = 90°, and sin(2θ) IS 1 after all. So you did the correct thing by setting it to 1.

    You understand why your first method is incorrect, right?
     
  6. Feb 11, 2013 #5
    Yes, stupidly I didn't realize that the first equation was taking acceleration in the x direction, not the y direction. Then, that cannot be used in the y direction (the direction that gravity is) :p.

    Also, do you know the difference between conservative and non-conservitive forces? I think that conservative forces and saved and can be used again. Like gravity, if you lift something up, the energy is conserved (like mgh) and friction, for example, the energy is conserved just not in a way that you can get back. Is my thinking correct on this?
     
  7. Feb 11, 2013 #6

    cepheid

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    Yeah, that's the basic idea. Slightly more formally, a conservative force is one for which the work done by the force along a path is independent of the path. It depends only on the starting and ending positions. This is the case for gravity. The work done by gravity in moving from point a to point b doesn't depend on what path you take between point a and point b. It only depends on the difference in height between points a and b. Mathematically, a force with this property will be the gradient (derivative) of a potential energy function. In one dimension this is expressed like this:$$\vec{F}(x) = -\frac{dU(x)}{dx}\hat{x}$$

    where U(x) is the potential energy as a function of position x. What this equation says is that a conservative force has an associated potential energy function. Its magnitude is equal to the rate of change of that potential energy, and it points in the direction of maximum decrease in that potential energy function.
     
  8. Feb 11, 2013 #7
    This is a calculus-based class, but my professor doesn't use calculus. I understand that so much better than his basic explanations. I don't know if he's afraid to use calculus or what, but I find that physics is easier using calculus (at least so far :p).

    Thanks for the help.
     
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