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Vertical Projectile Motion Problem

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball is projected vertically upward with a velocity of 30m/s. It strikes the ground after 8's.

    1.1) Determine the height of the building
    1.2) Find the height of the ball relative to the ground as well as its velocity at t=3 and t=5
    1.3) At what time will its velocity be 25m/s, downward?

    2. Relevant equations

    equations of motion

    3. The attempt at a solution

    1.1) Okay, I worked out the total time to go up and got 3's and then doubled it for coming down and got 6's (edge of the building). 8-6=2's coming down from top of the building.
    delta y=vi*t+0.5*a*t^2
    height of building = 79.6m

    1.2) at t=3's, v=0
    delta y = 45.9m
    Relative to ground:
    79.6+45.9=125.5m

    at t=5's
    vf = vi+at
    = 0+(9.8)*5
    = 49m/s - 30m/s
    = 19m/s

    delta y = 19.6m+79.6
    = 99.2m relative to the ground

    1.3) vf = vi+at
    t = 2.55+3
    = 5.55's

    Could someone please check?
     
  2. jcsd
  3. May 11, 2013 #2

    CWatters

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    I assume it was thrown from the top of the building..

    For 1) I used

    S = ut + 0.5at^2

    = 30*8 + 0.5(-9.8)82
    = 240 - 313.6
    = -73.6m

    If you assume g = 10m/s/s then you get

    = 240 - 320
    = - 80m

    -ve because it lands that distance below the height at which it was thrown from.
     
  4. May 11, 2013 #3

    CWatters

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    For 2)
    t=3
    S = ut + 0.5at^2
    S = 30*3 + 0.5(-9.8)32
    S = 90-44.1
    S = 45.9m (above the top of the building)

    Height above ground = 45.9 + 73.6 = 119.5m

    Repeat with t=5

    Edit: Gives 27.5 + 73.6 = 101.1m above ground.
     
    Last edited: May 11, 2013
  5. May 11, 2013 #4

    CWatters

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    For 3)

    v = u + at

    t = (v - u)/a

    v=-25

    t = (-25 - 30)/-9.8

    = 5.6 seconds
     
  6. May 11, 2013 #5

    CWatters

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    So all looks about right.
     
  7. May 11, 2013 #6
    Why did you use t=8?, because wouldnt that include the height that it was thrown up at?, they wanted the height of the building
     
  8. May 11, 2013 #7

    haruspex

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    The equation used tells you the height relative to the starting point after t seconds. It doesn't matter what route it takes to get there. E.g. note that t=0 and t=60/9.8 both give s=0.
     
  9. May 12, 2013 #8
    It took 3 seconds to reach max height after it was thrown and another 3 seconds to come down to its starting point(edge of building) so wouldnt you have to use 8-6= 2's for the time?, sorry just a little confused
     
  10. May 12, 2013 #9

    haruspex

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    Yes, but it's moving the opposite direction then. You seem to want to break it into stages, but there's no need. You can treat it as one smooth movement - in 8 seconds it will rise, then fall, and reach a point 73.6m below the start point. The quadratic equation encapsulates all of that.
     
  11. May 12, 2013 #10
    Think that cleared things up, thanks
     
  12. May 12, 2013 #11

    CWatters

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    Looks like you already got it but..

    t=0 is "a" time that it is it is at the top of the building
    t=8 is the time it is at the bottom of the building (eg the ground)

    so the displacement S is the height of the building.

    As haruspex said. The equation works for any point on the path even when that's below the starting point.

    On a seperate but related issue.. It's worth remembering that some problems have two possible solutions. For example a ball thrown up from the ground at an angle so that it lands on the roof of a building would have two solutions for the time of flight to height h. One on the way up and the other on the way down.
     
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