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Walking on the moon

  1. Jan 31, 2008 #1
    During a walk on the Moon, an astronaut accidentally drops his camera over a 14.7 m cliff. It leaves his hands with zero speed, and after 2.2 s it has attained a velocity of 3.3 m/s downward. How far has the camera fallen after 4.3 s?

    I am so lost! I have no idea what to do! I would appreciate any help!
     
  2. jcsd
  3. Jan 31, 2008 #2
    First you have to calculate (or search) the gravity acceleration on the Moon:

    g = G.M/R^2 (G: gravitational constant, M, R are the mass and radius of the Moon)

    Then everything is straight forward.
     
  4. Jan 31, 2008 #3
    then would i just multiply that answer by 4.3 for the time?
     
  5. Jan 31, 2008 #4
    With the information he was given, the following formula is better:

    http://en.wikipedia.org/wiki/Torricelli's_equation

    Vy^2 = Vy0^2 + 2aD

    Where: Vy is the final speed, Vy0 is the initial speed, a is the acceleration, D is displacement. a is the only unknown, so this should be fairly simple.
     
  6. Jan 31, 2008 #5
    I am trying to find how far it has fallen. I assume the final speed is the 3.3m/s stated in the problem and the initial is 0m/s. The acceleration on the moon that I figured out from this equation was 1.6256. So should I solve for the displacement? I am so frustrated. I have been working on this since 6 o'clock and I only have half the assignment done.
     
  7. Jan 31, 2008 #6
    Woops, I'm sorry, I thought you had the displacement from the first 2.2s...

    With the only known information being the variation of speed and time, you should use:

    Vy = Vy0 + a.t

    Where t is time. With the moon gravity obtained (which must be negative, make sure you get the signs right, and I got -1.5m/s), you have to use another equation:

    x = x0 + v0.t + 1/2.a.t^2

    Where x0 and x are initial and final position, respectively. You should be able to do this now. Now the exercise asks for the position at 4.3s. Solve for x, and make x0 = 14.7m.
     
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