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Linear Motion

  1. Nov 26, 2006 #1
    Neglecting air resistance, calculate the position (in meters) and velocity
    (in m/s) of a golf ball that is dropped from the empire state building after 1.0s, 2.0s, and 3.0 s with the release point.:surprised
     
  2. jcsd
  3. Nov 26, 2006 #2

    Hootenanny

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    One is expected to some some effort in answering the question...
     
  4. Nov 26, 2006 #3
    all i need is the equation to solve this problem...... not the answer...

    i'm partially leaning toward, v=delta(d)/delta(t) a=delta(v)/delta(t)
     
  5. Nov 26, 2006 #4

    Hootenanny

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    After the ball is release, what force(s) are acting on the ball?
     
  6. Nov 26, 2006 #5
    9.8m/s, which is gravity
     
    Last edited: Nov 26, 2006
  7. Nov 26, 2006 #6

    Hootenanny

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    That is an acceleration, not a force, but it was where I was leading to so no worries. Now, if an object starts at rest and accelerates at 9.8m/s2 (note the squared as opposed to your units) how fast will be be travelling after one second?
     
  8. Nov 26, 2006 #7
    9.8meters after 1 second
    then is it doubled and tripled after 2 and 3 seconds
     
  9. Nov 26, 2006 #8

    Hootenanny

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    I said how fast not how far.
     
  10. Nov 26, 2006 #9
    how fast.....9.8m/s

    this is where im stuck i dont know....... terminal velocity im guessing

    so possibly 75mi/hr after 5 seconds
     
    Last edited: Nov 26, 2006
  11. Nov 26, 2006 #10

    Hootenanny

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    Yes, that is correct. Have you met kinematic equations before? If not are you comfortable with energy conservation?
     
  12. Nov 26, 2006 #11
    no i do not know kinematic equations, but i do know a little about energy conservation...


    note: i have only been familiar with physics for about a week now.


    0.5m/s after 1.0seconds.... possibly
     
  13. Nov 26, 2006 #12
    i got 0.5m/s by solving for acceleration by dividing the change in velocity/over the time taken
     
  14. Nov 26, 2006 #13
    i crunched a few numbers and im coming up with this for the answer is this correct:

    after 1.0seconds its speed is 0.5m/s
    after 2.0seconds its speed is 1.0m/s
    after 3.0seconds its speed is 1.5m/s
     
  15. Nov 26, 2006 #14

    Hootenanny

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    What are you doing? You have given me the correct answer in one of your above posts;

    after 1.0 second speed = 9.8m/s
    after 2.0 seconds speed = 2 x 9.8 m/s
    after 3.0 seconds speed = 3 x 9.8 m/s

    Do you follow?
     
  16. Nov 26, 2006 #15
    so all i have to do to get the answer is to multiply 9.8 by two and 3

    there has to be more than that


    im really trying to understand
     
  17. Nov 26, 2006 #16
    there is not more than that--you are over complicating the simple concept of acceleration. 1m/s^2 means one meter per second PER SECOND. which means that every second it is going one meter per second faster. so 9.8 m/(ss) means 9.8 after one second, 9.8*2 after two and so on.
    do you know any calculus?
     
  18. Nov 26, 2006 #17
    Yeah...this is an incredibly easy physics question , you're stressing too much over it.
     
  19. Nov 26, 2006 #18
    acceleration = velocity/time
    velocity = acceleration * time
    Thats all there is to the velocity part. Because you are neglecting air resistance, the ball will continue to accelerate infinitely with no terminal velocity.
    Distance is a little trickier, you need the kinematic d=vo+.5at^2, which simplifies to 1/2at^2 in your case. When you do calculus, youll realize that this relationship between distance and acceleration is an anti-derivitive and theres a reason for that. But for now, just plug and chug
     
  20. Nov 26, 2006 #19
    ok i think then i got the answer to you the prob...

    after 1.0seconds its speed is 9.8m/s
    after 2.0seconds its speed is 19.6m/s
    after 3.0seconds its speed is 29.4m/s


    and i haven't taken calculus yet just algebra 2 I'm a freshman in high school.
     
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