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Why doesn't my result yield 9.8 m/s^2?

  1. Feb 15, 2014 #1
    A ball falling from rest is located 45meters below its starting point 3.0s later. Assuming its acceleration is uniform, what is it's value? (air resistance is negligible)

    Relevant Equations - Equations for uniformly accelerated motion
    Δd=(v1 x Δt)+(a x Δt^2)/2

    Attempt To solve:
    v1= 0
    Δd=45m
    Δt=3.0s
    a=?

    a=(2Δd/Δt^2)-(2v1Δt/Δt^2)
    a= ((2 x 45)/9)-((0 x 3)/9)
    a=10m/s^2

    If a ball is falling from rest(0 initial velocity) shouldn't it's acceleration be 9.8 seconds since gravity is constant? Is my equation wrong? Is my idea of time-acceleration misplaced?
     
  2. jcsd
  3. Feb 15, 2014 #2

    gneill

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    Staff: Mentor

    The given values result in the acceleration that you calculated. Your calculation is okay.

    In introductory classes g is often approximated to be 10 m/s2 because it simplifies numerical calculations while still getting across the concepts being taught.
     
  4. Feb 15, 2014 #3

    AlephZero

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    Science Advisor
    Homework Helper

    The question doesn't say the ball was near the surface of the earth. Don't make unjustified assumptions! :smile:
     
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