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(101)Acceleration due to gravity

  1. Feb 17, 2008 #1
    [SOLVED] (101)Acceleration due to gravity

    Ok, this should be really strait forward. Its only problem C that I am not getting correct.

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

    An object is observed to fall from a bridge, striking the water below 3.36 s later.
    (a) With what speed did it strike the water?

    (b) What was the average speed during the fall?

    (c) How high is the bridge?

    2. Relevant equations
    a) 9.8*3.36 = 32.928 m/s - correct

    b) Vi+Vf/2 = 16.464 m/s - correct

    c) ?

    3. The attempt at a solution

    Ok, acc due to gravity = 9.80m/s^2

    So, in the first second, the object fell 9.8m. In the second second, it fell 19.6m. t=3 it fell 29.4m. t=3.36 it fell 32.928m.

    Adding all of these would give me the total distance fallen (height of the bridge). The answer 91.728m is incorrect.

    Am I attacking this the wrong way?

    Thank you!
  2. jcsd
  3. Feb 17, 2008 #2
    There is a formula for calculating the distance travelled due to acceleration.... might want to give that a try
  4. Feb 17, 2008 #3
    why not use x=ut+1/2at^2, where x is distance, u is initial velocity, a is acceleration and t is time.
  5. Feb 17, 2008 #4
    Use the formula [tex]d = v_0 t + (1/2) a t^2[/tex]

    You know that initial velocity is 0. You know the time and you know the acceleration.
  6. Feb 17, 2008 #5
    >So, in the first second, the object fell 9.8m. In the second second, it fell 19.6m. t=3 it fell 29.4m. t=3.36 it fell 32.928m.

    If the object travels for 1 second at 9.8m/s, then it will travel 9.8m. Unfortunately, the speed in this case is not constant but increasing.
  7. Feb 17, 2008 #6
    I solved the problem with the above formula.

    I was saying that it fell at 9.8m/s per second, thus it fell 9.8m in the first second, and then 19.6 meters in the second second, meaning it had fallen a total of 28.6m at the end of 2 seconds. I dont understand why this approach doesnt work, but all is well with the d=Vot+ (1/2)at^2 formula.
  8. Feb 17, 2008 #7
    I think your approach ignores the fact that velocity is changing (Because there is acceleration) and it may be ignoring any initial velocity the object has already attained in the previous seconds.
  9. Feb 17, 2008 #8
    No, it fell less than 9.8m in the first second because it is only travelling at 9.8m/s for an instant. It would only fall 9.8m in one second if it were actually travelling at 9.8m/s for the whole second, but it (presumably) starts with zero velocity. You actually have to take the area under the velocity-time graph, so for the first second it travels .5*(9.8-0) m or 4.5m (and for constant acceleration, this turns out to be average velocity multiplied by time)
  10. Feb 18, 2008 #9
    That makes perfect sense. Thank you all for the help!
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