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1d kinematics, falling object

  1. Oct 4, 2005 #1
    Please tell me if I have done this correctly.

    A ball is thrown upward from the ground with an initial velocity of 10 m/s. Assume all digits are significant.

    a. How long before the ball reaches its maximum height?

    t = (0m/s - 10m/s) / -9.80m/s2 t = 1.0s

    b. What is the ball's maximum height?

    x = 0m + .5(10m/s + 0m/s) x = 5.0 m

    c. What is the velocity of the ball when it returns to the ground?

    v = 0m/s + -9.80m/s2 * 1.0s v = -9.8 m/s
     
  2. jcsd
  3. Oct 4, 2005 #2

    Päällikkö

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    What have you done in b?

    Never use rounded values (in the calculator), just give them as answers.
    b and c are wrong due to earlier roundings.
     
  4. Oct 4, 2005 #3
    We were instructed to round after each section of the problem. In b I used the formula x = x0 + 1/2(v0 + v)t. x0 is x naught and so on.
     
  5. Oct 4, 2005 #4

    Päällikkö

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    You are supposed to round the answer, but I've always been instucted to use unrounded values in the calculator.

    Anyways, the formula you used is wrong.

    The correct one (works for constant acceleration):
    [tex]x=x_0+v_0t+\frac{1}{2}at^2[/tex]
     
  6. Oct 4, 2005 #5
    x = 0m/s + (10m/s * 1.0 s) + (-9.80m/s2 * 1.0s^2)/2 = 5.1 m

    Ok, is this right?
     
  7. Oct 4, 2005 #6

    Päällikkö

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    Almost correct, the unit of x0 is not m/s, but m (you have x = 0m/s ...).
    It doesn't effect the result though, so: Yes, it's correct.
     
  8. Oct 4, 2005 #7
    It is better to write clean, nice and exact calculations.
    In some places they take off points in an exam if you don't do so .
    It aslo better to slove the question step by step and in a general form and insert numbers only at the final stage.
    Here is how I would slove this question:
    a.
    0) [tex]V_0=10m/s[/tex]

    [tex]V=0m/s[/tex]

    [tex]a=-9.8m/s^2[/tex]

    1) [tex]V=V_0+at[/tex]

    2) [tex]0=10-9.8t[/tex]

    3) [tex]10=9.8t[/tex]

    4) [tex]t=\frac{10}{9.8}=1.02...s[/tex]

    b.

    1) [tex]X=X_0+V_0t+\frac{1}{2}at^2[/tex]

    2) [tex]X=0+10*1.02+\frac{1}{2}*-9.8*1.02^2[/tex]

    3) [tex]X=12-5.09[/tex]

    4) [tex]X=6.91m[/tex]

    c. I think it is not hard find out (even without calculations)that the answer is -10m/s, try to figure out why by yourself (it is really easy).

    A little tip: Always make a drawing.
     
    Last edited: Oct 5, 2005
  9. Oct 4, 2005 #8

    Päällikkö

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    What did you, Alkhimey, do in b 3) ?
     
  10. Oct 4, 2005 #9
    Thank you everyone for your help. I know it's a simple problem. My instructor only gave me 2 points out of 20 for the answer, and I'm fairly certain that isn't reasonable. I've suspected that he was incompetent, but that's a different matter. I decided it would be best to take the course at a later date with a better instructor.
     
  11. Oct 5, 2005 #10
    [tex]10*1.02=12[/tex]
    [tex]1/2*-9.8*1.02^2=5.09[/tex]
    Therefore:
    [tex]X=12-5.09[/tex]
     
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