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Acceleration: im just confused

  1. Nov 15, 2004 #1

    DB

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    i'm new to physics and im just trying to figure out the force (Newtons) it would take to throw a baseball 100 mph (not concerning the forces of gravity acting upon it just the basic force you would need to throw it at that speed). i have already figured out that 100 mph is equal to a velocity of 44.704 m/s, knowing that the distance from the pitchers mound to home plate is 18.4404 meters and that it would take 0.4125 seconds to reach home plate. and i know the mass of a baseball is about .145 kg. so im trying to fill in Force = Mass x Acceleration. But i guess im doing this wrong because after aplying the formula a = m/s2 (squared) i get a = 108.37333334 m/s/s.

    I just dont understand how the acceration per second squared is larger than the velocity per second if acceleration is the rate of change of velocity. did i use the wrong formulas? please help, thanks
     
    Last edited: Nov 15, 2004
  2. jcsd
  3. Nov 15, 2004 #2

    ek

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    Don't square the time.

    a=d/t

    Edit: d/t? Wow....that's bad.

    :eek:

    V/T
     
    Last edited: Nov 16, 2004
  4. Nov 15, 2004 #3

    cepheid

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    Ok...first things first. This is not a formula for acceleration. The equation is not correct. It is true, that acceleration is measured in units of metres per second, per second i.e. m/s2. But if you want to calculate the acceleration, you should use Newton's second law. In this case, I don't think you have enough information to do so, since the force is an unknown as well. From what you have shown, it is not clear how you got your value for a. Could you please show your work in more detail? Thanks. :smile:

    EDIT...wait! Hold on...you changed the problem on me! That's cheating! lol.

    I'm still not sure what the problem is asking for.

    ek...a = d/t ??? That's not even dimensionally consistent.
     
    Last edited: Nov 15, 2004
  5. Nov 15, 2004 #4

    ek

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    He is doing this:

    18m/(.4s)2
     
  6. Nov 15, 2004 #5
    DB...
    Let me show you an example:
    [tex](1/2)^2=1/4[/tex]
    Answer these questions now
    1)any number (n) where 0<n<1, squared (like in the example) results in what magnitude change (larger of smaller)?

    Now, to go along with my example above, let me show you this: [tex]V=m/s[/tex] so... if the meters =10 and the seconds = .5, you get [tex]10/.5=V=20m/s[/tex]

    Knowing this, and keeping the same meters and seconds, we take acceleration, or [tex]a=m/s^2[/tex] so now we have (knowing that seconds^2=[tex].5s*.5s=.25s[/tex]) we get...[tex]10/.25=a=40m/s^2[/tex]

    As you can see, your acceleration is a lot larger than your velocity, because it has a time component (t) where 0<t<1.

    Now knowing this and knowing you have a time component less than 1 but greater than 0, (.4125 s) your problem with your problem (haha, sorry, I couldn't resist :rofl: ) should make sense.

    Cheers,
    Paden Roder
     
  7. Nov 15, 2004 #6
    Force, mass, acceleration

    To get the force, you need to know several things. Force acts on a mass to produce acceleration, according to N's 2nd Law: F=ma. So you'll need to know the mass of the baseball. Where does the force come from? Here, it is the pitcher who gets the ball up to speed. The acceleration only happens when the pitcher is pushing on the ball; after he (or she) lets go, the ball is on its own and maybe slows down by air resistance, so the distance from mound to home plate is irrelevant.
    You can estimate things here, to get a rough idea. Use the formula v^2=2ax to find a [a = v^2/(2*x)], v is 44 m/s, and x is the distance the pitcher pushes it, say twice the length of his arm? maybe 1.5m?
    Now search the web to get the mass of a baseball! check http://baseball-almanac.com/rule1.shtml, rule 1.09.
    With a and m known (convert ounces to kilograms), you can use F = ma to get the force (in newtons).
     
    Last edited: Nov 15, 2004
  8. Nov 15, 2004 #7

    DB

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    thanks alot

    thanks alot guys, i know the mass is about .0145 kg lol but im think now that acceleration is equal to d/t which would make it distance in velecity 1 - distance in velocity 2? a = dv1-dv2/t ???? thanks for the replys
     
  9. Nov 16, 2004 #8

    cepheid

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    1. What is the question asking? For the force required to accelerate the ball to 100mph? Is that it? If so...it depends on how long the force acts...which I don't see how we have any way of determining.

    2. I still don't understand any of this stuff other people are posting...where do 'formulas' like v = m/s, and a = m/s2 come from? It seems like nonsense. v = d/t, if the velocity is constant. a = (v2 - v1)/t...on average, or for constant acceleration. I'm not sure how these formulas are relevant here.

    3. Acceleration is NOT d/t. As I said before...check your units...what do you get when you divide distance by time? Metres per second. Are these units of acceleration? No. They are units of velocity. v = d/t, for constant velocity.

    :confused: (totally not understanding either what the question being asked is, or where people are getting their responses from).

    edit...except kmcf...he's making sense.
     
    Last edited: Nov 16, 2004
  10. Nov 16, 2004 #9

    ek

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    Bah...brain fart. I think you knew what I mean but I guess when I'm trying to help out a young'in I better not make such dumb mistakes.

    :tongue2:

    Sorry if my idiocy caused you any trouble DB.
     
  11. Nov 16, 2004 #10

    Integral

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    To find the force applied by the pitcher to the ball you will have to analyse the motions made in pitching the ball. You need to know the time from the instant the ball is at the rear most motion of the pitch till it leaves the pitchers hand. That time combined with the the velocity of the ball after it has left the pitchers hand will determin the acceleration and therefore the force applied.

    You have:
    Initial Velocity [itex] v_0 = 0 [/itex]

    Final Velocity [itex] V_f = 44 m/s [/itex]

    Forward motion time [itex] T_m [/itex]

    so

    [tex] a = \frac {V_f - V_0} {T_m} = \frac {V_f} {T_m}[/tex]
     
    Last edited: Nov 16, 2004
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