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Highest point of trajectory

  1. Jun 17, 2005 #1
    object thrown up with initial velocity of 15 m/s and horizontal velocity of 18 m/s, how much time is required to reach the highest point of trajectory?
    Vox = 18
    y = yo + vosint + 1/2gt^2
     
  2. jcsd
  3. Jun 17, 2005 #2

    Doc Al

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    Staff: Mentor

    I assume you mean that the initial vertical component of velocity is 15 m/s.

    What's the acceleration of the object? And what's the meaning of acceleration?
     
  4. Jun 17, 2005 #3
    a = 9.8 ??
     
  5. Jun 18, 2005 #4

    Doc Al

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    The acceleration due to gravity is 9.8 m/s^2 downward. Now what's the relationship between speed, acceleration, and time?
     
  6. Jun 18, 2005 #5
    1st it can be assumed that y0 =0[origin]

    y=V0 sin(theta) t - 1/2 gt^2... g's sign will be -ve here due to upward motion...

    [i cant understand why sint is written...it should be some angle not time]

    at highest point dy/dt=0 and

    0= V0 sin(theta) -gt.......[put t=T/2] wheret is total time of flight-----iii

    V0x=V0 cos (theta)=18---------i
    15cos(theta)=18

    find out theta
    put it in iii
    T= 2V0 sin(theta)/g
     
  7. Jun 18, 2005 #6

    Doc Al

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    All you need consider is the vertical component of the motion:
    [tex]v_f = v_i + at[/tex]
    (which is merely a restatement of what acceleration means)
     
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