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Free fall of ball and acceleration

  1. Sep 6, 2003 #1
    I suppose I am having trouble with this problem because there are no numbers given. Here is the problem:

    A ball is thrown down vertically with an initial speed of Vo from a height of h.

    a) What is its speed just before it strikes the ground?

    Speed is just the magnitude of instantaneous velocity, right? So I just used the equations for velocity:

    Vf =?
    V^2 = Vo^2 + 2a(Y-Yo)

    Since I'm dealing with free fall accel, a = -9.8 m/s^2. In addition, Y-Yo = h. So,

    V^2 = Vo^2 + 2(-9.8)h --> V^2 = Vo^2 -19.6h

    Leaving me with:

    V = (Vo^2 -19.6h)^(1/2)

    b) How long does the ball take to reach the ground?

    I used: Y-Yo = Vot + (1/2)at^2

    and using the quadratic equation, I got:
    ** i used the "+" version bec t cannot be negative

    t = [-Vo + (Vo^2 - 19.6)^(1/2)]/-9.8

    c) and d) are the same as a) and b) except this time the ball is thrown upward. I also have to figure out if the values for c and d are less than, greater than, or the same as the values in a and b.
    <b>c)</b> I used:

    Y - Yo = (1/2)(Vo + V)t
    h = (1/2)(Vo + V)t
    V = (2h/t) - Vo

    I think the values will be the same.

    d) For this part, I used the V found in c)

    V = (X1 - X)/t1 - t --> V = h/t --> t = h/V

    this is the time just to fall from the maximum point the ball reaches, so the total time to fall to the ground would be:

    2t = 2(h/V)

    Because the time is doubled, I think this value will be greater than the value in b)

    Were my assumptions correct?
  2. jcsd
  3. Sep 6, 2003 #2


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    Science Advisor

    The formulas you are using are correct.

    In part (b), you appear to have left out the "h" in 19.6h

    In part (c), the answer should be the same as (a). If you throw the ball up at speed v0, when it passes the level from which it was thrown, on its way down, it will now have speed v0 downward (neglecting air resistance) just as in problem c.

    "Because the time is doubled, I think this value will be greater than the value in b)"
    Yes, of course, the time in part d will be greater. After you have thrown the ball up, it will (as you say) take a specific time to read it's maximum, the same time to reach level "h" again and THEN the same time as in (a) to hit the ground. In fact you can find that time more eaily by using the same formula as before (with y-y0= 0 instead of h) to calculate the time the ball will take to come back to height h, then add the answer from part (b).
  4. Sep 6, 2003 #3


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    (a) Mistake! You have to keep in mind the DIRECTION of the acceleration. Objects acccelerate downwards. You are also throwing the ball... downwards. And the ball is travelling... you guessed it, downwards. Therefore, v, a and h should all have the same sign, giving V^2 + 2 * 9.8 * h.

    (b) See above.

    (c) Yep. But your derivation is a bit wrong. Look back at the equation you used for (a), and ask yourself - what difference would changing the sign of Vo make?

    (d) Uh... I am confused as to your method...
  5. Sep 11, 2003 #4
    correction to problem = additional question

    I was rereading the problem in the book and I seemed to have missed one of the words. For part c and d the question is What would be the answers to a and b if the ball was thrown upward from the same height and with the same initial speed?

    With this in mind, I came up with the following answers:

    c) V = (Vo^2 - 19.6(u+h))^(1/2)
    where h is the original height and y is the the extra distance the ball traveled when thrown up from that initial height, h. With that in mind, I believed that this value would be slightly larger than the original answer for a becuase of adding u to h.

    d) t = [Vo +- (Vo^2 - 19.6(u+h))^(1/2)]/9.8

    I also thought that this value would be slightly higher than the original value found in b because of the additional u added to the height, h. How would I know whether to choose + or the - value for computing the time, t?

    With this new information, are my answers correct?
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