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Free fall problem, Help?

  1. Feb 9, 2010 #1
    Free fall problem, Help!?

    1. The problem statement, all variables and given/known data

    Hi, Im having trouble figuring out this problem:

    A ball is thrown directly upward with an initial velocity of 10 m/s. If the ball starts at an initial height of 2.3 m, how long is the ball in the air? Ignore air drag.

    2. Relevant equations

    The equation that I used to solve this was:

    delta-y= (vi)t-1/2gt²

    3. The attempt at a solution

    I plugged in my numbers so i got:

    -2.3= 10t-1/2(9.8)t²
    -2.3=10t-(4.9)t²
    4.9t²-10t-2.3=0

    then i did the pythagorean theorem(but I dont really know how to type it on the computer.. but i will try)

    -10 ± √((-10²)-4(4.9)(-2.3))/ 2(4.9)

    -10 ± √(100-45.08)/2(4.9)

    -10 ± √(54.92)/ 9.8

    (-10 ± 7.41)/ 9.8

    1.77 s or -1.77 s


    So I put the answer 1.77 s and it was wrong. i put 1.8s and it was wrong. I used 1.02 and it was wrong. I used 2.04 and it was wrong. I have no idea what i am doing wrong and I no longer have any more submissions for my answer (which means i oficially have a zero for that question). But i really need to know how to solve this problem! please help me!

    thanks!
     
  2. jcsd
  3. Feb 9, 2010 #2
    Re: Free fall problem, Help!?

    You just made a couple mistakes with your sign:

    You have
    4.9t²-10t-2.3=0 which is correct but when you use the quadratic formula you have (it's not the Pythagorean btw)

    -10 ± √((-10²)-4(4.9)(-2.3))/ 2(4.9)

    but it should be

    (10 +/- sqrt[(-10)^2 - 4(4.9)(-2.3)])/(2(4.9))

    which is

    ((10 +/- sqrt[100 + 4(4.9)(2.3)])/(2(4.9))


    that should give you the correct answer
     
  4. Feb 9, 2010 #3
    Re: Free fall problem, Help!?

    a=4.9
    b=-10 (not 10!)
    c=-2.3
     
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