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Problems with acceleration due to gravity, and terminal speed

  1. Sep 11, 2006 #1
    My first question:

    1. A baseball pitcher throws a ball vertically upward and catches it at the same level 4.2s later.

    a) With what velocity did the pitcher throw the ball?
    b) How high did the ball rise?

    (my work, which is wrong according to the book)


    [itex]v_2 = v_1 + (a)\Delta(t)
    = 0m/s - (-9.80m/s^2)(4.2s)
    = 41m/s [up][/itex]

    The book says it's 21m/s [up], but I don't really see why, so could someone please tell me what I did wrong, or if I'm actually correct, while the book is wrong(the books have been wrong several times before).


    I haven't started this yet, because I think I need the initial veolocity in order to calculate the displacement. Would I be correct to use this equation?

    [itex]\Delta(d) = v_1\Delta(t) + 1/2a\Delta(t)^2[/itex]

    Now for my second question:

    There are many well-documented cases of people falling from tremendous heights without parachutes and surviving. The record is held by a Russian who fell from an astounding 7500m. The chances of survival depend on the "deceleration distance" at the time of landing. Why is a fall from a height of 7500m no motr dangerous than one from half that height? How can the deceleration distance upon landing be maxized?

    I think I got the first part, because of terminal speed, there wouldn't be a difference in between the two speeds acheived from the falls(assuming the fall from half that height achieves terminal speed). But the second part has me stuck, the question doesn't really explain what the deceleration distance is, so I don't know how to maximize it, or much about it all, for that matter.
    Last edited: Sep 11, 2006
  2. jcsd
  3. Sep 11, 2006 #2


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    a) x= -(9.8/2)t2+ v0t. If x= 0, t= 0 is one solution. You are told that the other is 4.2 s. What is v0.

    b) v(t)= -9.8t+ v0. The ball will continue to go up as long as the velocity> 0. It will be descending when the velocity< 0. The maximum occurs when v= 0. What value of t makes that true?
  4. Sep 11, 2006 #3
    So do you have any references for this well documented case?
  5. Sep 11, 2006 #4
    For a), I don't really know what you mean, do I sub in 4.2 for t, and then solve for vo? if so, what value of x do I use, zero?

    For b), I'm not really sure what you mean here, either, what's vo here?

    No, I just typed up the question word for word.
  6. Sep 12, 2006 #5
    Could someone please give me some more help with this?
  7. Sep 12, 2006 #6

    Doc Al

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    Your error is in using t = 4.2s. If it returns to his hand in 4.2s, how much time did it take to reach the highest point? (Using the right time, your work would be correct. You are mixing up v_1 and v_2 and signs a bit.)

    If you wish to use t = 4.2s, then realize that v_1 = - v_2.

    That equation will work fine. (You don't really need the initial speed if you start at the top of the motion.)
  8. Sep 12, 2006 #7
    Ok then, so what time do I use? Would I assume that the trip took half the time for the way up?

    Also, I was wondering something about this:

    Why would I use 4.2s at all, if it's not the correct time?
  9. Sep 12, 2006 #8

    Doc Al

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    Yes. It spends the same amount of time rising as falling.

    That is the correct time, for the entire "trip" of the ball. If you want to consider just half the trip, then use half the time. If you want to consider the entire trip, use the entire time. Depending upon what you are solving for, sometimes either way will work, sometimes you must use one or the other.

    If you analyze just half of the trip, then:
    v_1 = ?
    v_2 = 0
    v_2 = v_1 + at (t = 4.2s/2)

    If you analyze the full trip, then:
    v_1 = ?
    v_2 = - v_1 (when it comes down it's going at the same speed in the opposite direction)
    v_2 = v_1 + at (t = 4.2s)
  10. Sep 13, 2006 #9
    How do you know that the trip took the same amount of time both ways? This is probably pretty straight forward, I think I'm just forgetting something.
  11. Sep 13, 2006 #10

    Doc Al

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    By symmetry: The trip down is the reverse of the trip up.
  12. Sep 14, 2006 #11
    Symmetry? Is this related to the parabolic path the projectile follows?
  13. Sep 14, 2006 #12

    Doc Al

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    Yes, the parabolic path is symmetric for the same reason.

    More simply, I mean this: Toss a ball in the air. Whatever vertical speed it has going up, it will have that same speed going down when it reaches the same point from the top.
  14. Sep 14, 2006 #13
    But why exactly do you know this? I'd think that in order for you to know this, you'd need to know either the time of the trip down, or the final velocity downwards, or can on of these be estimated from a graph, or by some other means?
  15. Sep 14, 2006 #14

    Doc Al

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    All you need to know are the equations of motion. For example: The vertical speed is described by V = V0 + at. Let's call t = 0 to be the time when the ball is at the top of its motion, thus V0 = 0 and V = at. So, one second before it reaches the top (t = -1), the speed is V = +9.8 m/s^2; one second after it reaches the top (t = +1), the speed is V = -9.8 m/s^2.

    Position works the same way. The vertical position (with t = 0 & y = 0 at the top) is given by: y = 0.5at^2. So at t = -1 & t = +1 the position is the same: y = -4.9 m.
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