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Constant acceleration on Mars

  1. Jan 5, 2010 #1
    1. A tennis ball on Mars, where the acceleration due to gravity is .379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after being hit?



    2. v (y)= v (0y) + (a (y) * t)
    y=y (0) + v (0y)t + .5 a(y) t^2
    v^2 (y)= v^2 (0y) + 2a(y) (y-y(0))
    They're the equations of motion for constant acceleration




    3. Well, the acceleration of gravity on Mars is 3.7 m/s^2. The answers are (a) 33.5 m, and (b) 15.8 m/s. I'm not sure how they got these answers, though. Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Jan 5, 2010 #2

    Oddbio

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    Gold Member

    First of all, the default format in the homework section has ending bold marks for a reason. It's a bit unpleasant to read an entire bold paragraph.

    Just think it through, you have the equation you need. The problem talks about time and position, you have a function for the position y(t).
    Hint: say that your 8.5s = T then according to the problem y(0) = y(T)
     
  4. Jan 5, 2010 #3
    Sorry for the bad format, I'm still trying to get the hang of the forum.

    OK, I'm trying to understand the logic behind this. From what I gathered in the textbook, to get the highest point, just assume that the final velocity is 0, and figure it out from there, but the original velocity needs to be known, which isn't in this case. Am I at least on the right track? Sorry for the dumbness, but I'm just getting back into Physics after multiple years away.
     
  5. Jan 5, 2010 #4

    ideasrule

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    So use one of the equations to calculate the initial speed and use another to calculate height.
     
  6. Jan 5, 2010 #5

    Oddbio

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    Well you could do it in pieces like that where the final velocity is zero.
    In that case, what is the time for the ball to go up to it's highest point? (with no air resistance)
    You shouldn't need any equations for that.
     
  7. Jan 5, 2010 #6
    4.25?
     
  8. Jan 5, 2010 #7
    OK, I got it. I feel like a complete idiot, man, lol. The initial position (y 0) is 0, and the final position (y) is 0. I use the second equation to find the initial velocity. I then plug it into the final equation, and make the final velocity 0, and I get both of them. Wow, how stupid of me.
     
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