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Basic Physics Question please help

  1. Oct 10, 2006 #1
    2 seconds after being projected from ground level, a projectile is displaced
    40 m horizontally and 53 m vertically above its launch point. What are the horizontal and vertical components of the initial velocity of the projectile? At the instant the projective ahieves maximum height above ground level, how far is it displaced horizontally from the launch point?

    Any suggestions for that questions, i've already drawn a diagram.
     
  2. jcsd
  3. Oct 10, 2006 #2

    Doc Al

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    Start by writing equations that describe the horizontal and vertical displacement as a function of time.
     
  4. Oct 10, 2006 #3
    so would saying the following make any sense:

    the horizontal velocity is constant so the projectile is moving 40m/s right
    the vertical velocity changes 9.80 m/s downward because of gravity
     
  5. Oct 10, 2006 #4

    Doc Al

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    The horizontal velocity component is constant, but it's not 40 m/s.
    The vertical velocity component accelerates (due to gravity) at 9.8 m/s^2 downward; acceleration is the rate of change of velocity (m/s per s).
     
  6. Oct 10, 2006 #5
    how do i find the horizontal velocity component?
     
  7. Oct 10, 2006 #6

    Doc Al

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    Use the basic definition of velocity.
     
  8. Oct 10, 2006 #7
    can i do this:

    y = 1/2 * g* t^2
    y = 1/2 (9.8) (2)^2
    y = 19.6
     
  9. Oct 10, 2006 #8
    or is that only when an object is thrown directly upwards and falls directly downwards.
     
  10. Oct 10, 2006 #9
    alright, i know that v = d/t so i'd get v = 40m/2sec = 20 m/sec but do i use the horizontal distance or vertical distance ?????? and what does velocity give me???? is that the horizontal component?????
     
  11. Oct 10, 2006 #10

    Doc Al

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    Well... 40m is the horizontal distance... so that must be the horizontal component of velocity.
     
  12. Oct 10, 2006 #11
    now how do i find vertical component?
     
  13. Oct 10, 2006 #12

    Doc Al

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    Do what I suggested in post #2. What equation describes the vertical motion?
     
  14. Oct 10, 2006 #13
    y = y0 + v0y*t - 1/2*g*t^2 ?
     
  15. Oct 10, 2006 #14
    so:

    - initial position (y0) would be 0 m
    - initial velocity (v0) would be 0 m/s
    - time would be 2.0 sec
    - force of gravity 9.8 m/s^2

    so my answer would be -19.6 m/s??
     
  16. Oct 10, 2006 #15
    wait.. wouldn't gravity be -9.8 m/s^2 so that the answer would be 19.6 ???

    and i don't know what units it'd be
     
  17. Oct 10, 2006 #16
    what makes you think its initial velocity is 0
     
  18. Oct 10, 2006 #17
    its on the ground?
     
  19. Oct 10, 2006 #18
    The velocity you calculated is if you dropped it off a cliff (in which case its initial velocity would be zero). However, in this case it is projected upwards meaning it was given some initial velocity
     
  20. Oct 10, 2006 #19

    Doc Al

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    OK.
    v0 is what you are trying to find! Don't assume it's anything--certainly not zero. You know it moved up so v0 can't be zero!

    OK
    The acceleration due to gravity is -9.8 m/s^2.

    You left out one piece of data: the vertical displacement at time t.

    Solve for v0.
     
  21. Oct 10, 2006 #20

    take into consideration his equation has a built in negative so if he put g in as negative it would make the object fly upwards
     
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