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Free-fall of a ball

  1. Jan 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A red ball is thrown down with an initial speed of 1.3 m/s from a height of 26 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

    What is the speed of the red ball right before it hits the ground?

    2. Relevant equations
    x = 26 - 1.3t - 4.905t^2


    3. The attempt at a solution
    I get how to do this problem, but I wasn't able to get the correct answer the first time when I did it completely mathematically. Can someone check where I'm going wrong in this work?

    We need to find the time that the ball hits the ground --> set x = 0
    0 = 26 - 1.3t - 4.905t^2
    -26 = -1.3t - 4.905t^2
    -26 = (-1.3 - 4.905t)*t
    t = 0,
    -4.905t = -24.7
    t = 5.03 seconds

    I don't know where I'm going wrong to get that t = 5.03 seconds, when the answer is 2.17 seconds (by using a calculator and plotting the zeroes). Is it wrong to solve the quadratic like this when solving for the roots?
     
  2. jcsd
  3. Jan 19, 2016 #2

    Ray Vickson

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    Homework Helper

    If ##-26 = t(-1.3-4.905 t)## you cannot have ##t = 0##. If you set ##t = 0## you get right-hand-side = 0 (because 0 times anything = 0), but left-hand-side = -26, and 26 ≠ 0.
    So, yes, it is very wrong indeed to solve quadratic equations that way!

    Why don't you just use the familiar quadratic equation roots formula?
     
  4. Jan 19, 2016 #3
    Yeah I guess I just confused that approach with solving the time for when x = some number. Thanks for pointing that out!
     
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