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Determing Height To Which A Projectile Will Rise

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    At the Earth's surface, a projectile is launched straight up at a speed of 9.6 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.


    2. Relevant equations
    Conservation of mechanical energy [itex]K_i + U_i = K_f + U_f[/itex]


    3. The attempt at a solution
    I am quite confident that I chose the correct formula to solve this problem; however, I have a few questions regarding the formula itself.

    I know that [itex]K_i[/itex] and [itex]U_i[/itex] won't be zero, because the projectile has an initial velocity and it is above the surface of the earth. Would [itex]K_f[/itex] be zero, though; because we are considering the highest point the projectile will reach, and gravity will necessary reduce its speed to zero, corresponding to the highest point. Because [itex]K_f = 0[/itex], [itex]U_f[/itex] can't equal zero, right? What would the situation be like if it was negative, how about positive?
     
  2. jcsd
  3. Jan 28, 2013 #2

    PeterO

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    There seems to be a contradiction in the two pieces I have shaded red.

    Energy is a scalar - so is neither positive nor negative. That makes your final sentence rather curious.
     
  4. Jan 28, 2013 #3

    tms

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    Last time I checked, scalars had signs.
     
  5. Jan 28, 2013 #4

    tms

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    Only differences between potential energies have physical meaning, so the zero point is arbitrary; you can set the potential energy to zero at any convenient point.
    Well, would it?
     
  6. Jan 28, 2013 #5
    If had the knowledge to answer your question, then I wouldn't have asked my question in the first place.
     
  7. Jan 28, 2013 #6
    Yes, I suppose I can see how the ambiguity could arise. What I intended for the statement, "..it is above the surface of the earth," was that the projectile wasn't below the the surface, that is, underground.
     
  8. Jan 28, 2013 #7

    PeterO

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    Above the surface and below the surface represent only 2 of the 3 possibilities for the projectile's original position.

    The opening words referred to that 3rd possibility.
     
  9. Jan 28, 2013 #8

    tms

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    Surely you have enough knowledge by now to say what the kinetic energy of an object with zero velocity is.
     
  10. Jan 28, 2013 #9

    HallsofIvy

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    If you really don't know how to calculate kinetic energy, rather than using energy, you could solve this problem with the basic [itex]h(t)= h_0+ v_0t- (g/2)t^2[/itex].

    You can take [itex]h_0[/itex] to be 0 at the surface of the earth and you are given that [itex]v_0= 9.6[/itex]. Of course, g= 9.82 m/s^2, approximately.
     
  11. Jan 28, 2013 #10

    PeterO

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    Given the initial velocity of 9.6 km/s this mass is going to rise more than 4000 km. It may not be reasonable to assume that g has a constant value to that elevation.

    You probably need a potential energy formula like U = -GMm/R
     
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