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Constant Acceleration in 1-D

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data

    1. Rob throws a baseball upwards at 14.2 m/s. His friend, John, is sitting in a tree 4.5m above Rob.

    a. Calculate how long it will take to reach John.

    b. If John misses the ball as it moves upwards, how long will it take to reach John again.

    2. Relevant equations
    Δd=V(Δt) + 1/2 (a)(Δt)^2

    3. The attempt at a solution
    I have found the time using the quadratic equation for part a which is 0.36s, but I am stuck on part b. I do not understand how to do it. It seems like Δd is the same in both parts. How in hell do I find Δt for part b.
     
  2. jcsd
  3. Sep 21, 2015 #2

    haruspex

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    When you solve a quadratic, how many solutions do you get?
     
  4. Sep 21, 2015 #3
    Well, you get 2 roots, but one is deemed useless because it is a negative.
     
  5. Sep 21, 2015 #4

    billy_joule

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    If you set up your equation correctly you'll get two positive roots, one being the 0.36 s you've found, and another at some later time.
    Show your working and we'll see what went wrong.
     
  6. Sep 21, 2015 #5
    -14.2 ± √113.35
    -9.81

    Δt=0.36s
    Δt=-2.53s​
     
  7. Sep 21, 2015 #6

    billy_joule

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    Your quadratic equation is correct, you've made an error somewhere while solving it.
     
  8. Sep 22, 2015 #7

    haruspex

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    I'm guessing you made a sign error. Initial velocity is positive but acceleration is negative, so your -14.2 is the wrong sign. If you cannot spot your error, please post all steps.
     
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