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Homework Help: Acceleration due to gravity

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A pitcher throws a ball vertically upward and catches it at the same lever 4.2 seconds later.

    A) What velocity did the pitcher throw the ball?
    B) What distance did the ball travel?


    2. Relevant equations
    Vfy^2=Viy^2+2ay[tex]\Delta[/tex]y
    [tex]\Delta[/tex]y=Viy+0.5ay([tex]\Delta[/tex]t)^2


    3. The attempt at a solution

    we know
    [tex]\Delta[/tex]t=4.2s
    ay=+9.8m/s/s,[Up] -9.8m/s/s[Down]
    Viy=?
    Vfy=?
    [tex]\Delta[/tex]y=?

    Several attempts have failed lol. One was to take the balls maximum height as initial velocity=0 and try to figure out distance (y) and final velocity. epic fail.

    I think my main problem is trying to understand how to re arrange these equations, depending on the quantity i am missing. The book suggests the use also of the quadratic equation, and epends all of 4 sentances explaining it. could it be used here?

    One other question, in an attempt to better understand acceleration due to gravity.
    Since he is throwing the ball up, at whatever velocity he throws it, the distance up will be the same as the distance down, providing he catches it on the same level (obviously) and since the the force slowing it down on the way up, gravity, at a rate of 9.8m/s/s is also the force and rate speeding it up on the way down the final velocity should be the same as initial velocity, right? forgive me if this is an extremely stupid obvious question haha, all this math is makeing my head spin
     
  2. jcsd
  3. Mar 10, 2010 #2

    tiny-tim

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    Hi slpnsldr! :smile:

    (have a delta: ∆ and try using the X2 and X2 tags just above the Reply box :wink:)
    Use the second equation. :wink:

    What do you get? :smile:
     
  4. Mar 10, 2010 #3
    delta y represents change in y position right? If he throws it straight up and catches it 4.2 seconds later in the same spot, what is the change in y position? Once you have that cleared up in your head it shouldn't be a problem.
     
  5. Mar 10, 2010 #4
    tiny tim, if i use the second equation, i need to know the initial velocity. unless, i split the balls motion in half, and calculate for [down], using 0 for Vi. Which means i would have to also cut delta time in half

    delta y= 0(4.2)+0.5(-9.8)(2.1)
    = -10.29m [down]
     
  6. Mar 10, 2010 #5

    tiny-tim

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    Nooo :redface:

    s = vit + 1/2 at2

    you know s and a and t, and the question is asking for vi anyway. :wink:
     
  7. Mar 11, 2010 #6
    hmm i dont quite understand what you mean tiny tim... is s=∆y? i dont know the distance... and dont i need to know initial velocity to use the equation your suggesting.. sigh lol im not very good at this :frown:
     
  8. Mar 11, 2010 #7

    tiny-tim

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    Yes, you do …
    (i assume you mean "level" ?)

    … the distance is zero. :wink:

    (and yes, s = ∆y in this case)
     
  9. Mar 11, 2010 #8
    hmm i dont quite understand what you mean tiny tim... is s=∆y? i dont know the distance... and dont i need to know initial velocity to use the equation your suggesting.. sigh lol im not very good at this :frown:
     
  10. Mar 11, 2010 #9

    tiny-tim

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    The ball leaves the pitcher's hand, and returns to the hand.

    The total distance (or displacement ) ∆y, is therefore zero.

    So you can put s = 0 into your equation. :smile:
     
  11. Mar 11, 2010 #10
    You disregarded what i said good sir. read what i said again, once you understand what the change in distance is, you will understand the problem.
     
  12. Mar 11, 2010 #11
    Sorry dac! I think i do understand now.. but.. the question asks at what velocity did he throw the ball, so i still have to figure out the balls velocity as it moves up right? first though distance should be...
    s= vit + 1/2 at2
    =0(4.2)+0.5(9.8)(4.2)^2
    =86.436m

    ok, so now... i have the total distance that the ball has traveled.. and now i need to rearrange an equation? to find the initial velocity of ∆y[up] which should be 43.218m[up] (half of s)
     
  13. Mar 11, 2010 #12

    tiny-tim

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    Nooo!

    s is zero, not vi. :redface:
     
  14. Mar 11, 2010 #13
    hmm, so i need to rearrange?

    vi=s/t+1/2at2?
     
    Last edited: Mar 11, 2010
  15. Mar 12, 2010 #14

    tiny-tim

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    No!!

    The equation is correct … s = vit + 1/2 at2.

    But s in this case is zero,

    so in this case 0 = vit + 1/2 at2.
     
  16. Mar 12, 2010 #15
    so... 0=0(4.2s)+0.5(9.8)(4.2)2
    =172.872
     
  17. Mar 12, 2010 #16

    tiny-tim

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    No, vi is the initial speed of the ball, it is not 0.

    0 = vi(4.2s) + 0.5(-9.8)(4.2)2
     
  18. Mar 12, 2010 #17
    lol oh man I`m so lost!! I should take this moment to first of all thank you for your patience lol. and then pray it will last a bit longer!

    So, what im trying to find is the initial velocity. the velocity of the ball as it leaves the hand of the pitcher.

    i now realise that the final change in y position is 0. but I do not understand what we are doing with this equation

    0=Vi(4.2s)+0.5(-9.8)(4.2)2
     
  19. Mar 12, 2010 #18
    lol oh man I`m so lost!! I should take this moment to first of all thank you for your patience lol. and then pray it will last a bit longer!

    So, what im trying to find is the initial velocity. the velocity of the ball as it leaves the hand of the pitcher.

    i now realise that the final change in y position is 0. but I do not understand what we are doing with this equation

    0=Vi(4.2s)+0.5(-9.8)(4.2)^2
     
  20. Mar 12, 2010 #19

    tiny-tim

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    Obviously, you divide by 4.2 to get vi = 0.5(9.8)(4.2)

    Have you done calculus?​

    If so, the reason is that d2y/dt2 = a.

    Integrate once, and you get dy/dt = constant + at,

    Integrate again, and you get y = constant + (constant)t + at2/2.

    (the first constant is yi, the second constant is vi)
     
  21. Mar 12, 2010 #20
    wow that works.. Vi=0.5a(t)

    I haven't taken calculus.. I haven't taken much actually, I'm taking some DL courses now to brush it all up, thought i was doing fine trying to teach myself until this question came up.. I'm afraid i still dont completely understand. How did we jump from
    0=Vi(4.2s)+0.5(-9.8)(4.2)^2 to Vi=0.5a(t) what did we divide by 4.2? how did we know to do this? and how did the Vi get to the other side of the equal sign?
     
  22. Mar 12, 2010 #21
    but this is a formula i can use, my question is answered.. im just trying to understand it now haha
     
  23. Mar 12, 2010 #22

    tiny-tim

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    Because we had s = ut + 1/2 at2.

    We put s = 0, giving 0 = ut + 1/2 at2,

    then we moved the ut across the equal sign, but multiplied by -1, to give ut = -1/2 at2,

    and then we divided by t to give u = -1/2 at. :smile:
     
  24. Mar 12, 2010 #23
    Well cool!!! Thanks so much :)

    However!!! I have come across another problem in this evil evil question!!!

    Why, doesn't this equation work to figure out the distance the ball travels?

    [tex]\Delta[/tex]y =Vi[tex]\Delta[/tex]t + 1/2a([tex]\Delta[/tex]t)2

    When put in what we know i get the answer

    174.636
     
  25. Mar 12, 2010 #24

    ideasrule

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    That's because delta-y is displacement, not distance. You figured out that v=20.6 m/s. You know a=-9.8 m/s^2 and delta-t=4.2 s, so if you plug in the numbers, you get delta-y=0. That's exactly what was stated in the question.

    However, part b) is not looking for delta-y; it's looking for the total distance that the ball travels. You can still use that equation, but in a slightly modified way.
     
  26. Mar 13, 2010 #25

    tiny-tim

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    (just got up :zzz: …)

    In other words: find the half-way distance, and double it! :smile:
     
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