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How to find the final height of a moving object

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A startled armadillo leaps upward, rising 0.523 m in the first 0.206 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.523 m? (c) How much higher does it go?


    2. Relevant equations

    The five Kinematic equations

    3. The attempt at a solution

    I worked out A and B and got the following answers:

    (a) 3.54 m/s

    (b) 1.5 m/s

    I just can't figure out the last part, C, on how to find how much higher the armadillo goes. Can anyone help me out with this?
     
  2. jcsd
  3. Aug 30, 2009 #2
    I would use [tex]2a(x-x_0)=v^2-v_0^2[/tex]. Do you see how you can use this to solve (c)?
     
  4. Aug 31, 2009 #3
    I'm using WileyPlus and its not taking any of the answers I have come up with.

    I inserted all the variables into this modified equation to find the final x variable:

    X=(V^2-V0^2+2aX0)/2a

    Did I do the algebra wrong?

    I keep getting 1.047m.
     
  5. Aug 31, 2009 #4

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    That's wrong, and it looks like you are plugging in the wrong velocities here.

    For part (c), the armidillo's "initial" velocity is the 1.5 m/s it has at the 0.523 m height. It's "final" velocity is the velocity it has when it has reached it's maximum height (and that velocity is ____?)
     
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