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What is the projectiles time of flight?

  1. Sep 26, 2004 #1
    Havin a bit of trouble on this one, just because of the lack of information.

    A projectile is launched over level ground at 35 m/s at an angle of 40 degrees above the horizontal. What is the projectiles time of flight?

    Someone mind helpin me out? :smile:
     
  2. jcsd
  3. Sep 26, 2004 #2
    Ive figured out the velocity of the X and Y components but i dont know where to go from there...
     
  4. Sep 26, 2004 #3

    Tide

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    You can determine the vertical component of the velocity at t = 0 so you know that
    [tex]y = v_{y0}t - gt^2/2[/tex]
    from which you can determine when the projectile hits the ground.
     
  5. Sep 26, 2004 #4
    Consider only the vertical motion of the projectile. You have intitial and final vertical velocities, as well as the acceleration (g), so how to find time?
     
  6. Sep 26, 2004 #5
    Doesn't that involve finding vertical distance (y) first? The formula I was getting at allows you to solve this question in one step.
     
  7. Sep 26, 2004 #6
    hmm, sorry I still cant figure it out :confused: Does Vf = 0 in this situation?
     
  8. Sep 26, 2004 #7
    Well, think about the unidirectional motion upwards (forget the trip back down for now). What happens when the projectile reaches the summit of its trajectory?
     
  9. Sep 26, 2004 #8
    The Velocity is equal to 0. But how could that be if the object isnt just going up and down? Its always moving isnt it?
     
  10. Sep 26, 2004 #9

    robphy

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    "[itex]y[/itex]'' is the specific height when "[itex]t[/itex]'' is a specific time.
    So, if you are looking for the "time when it meets the ground" ([itex]t=t_\text{ground}[/itex]) then "[itex]y[/itex]'' is the height at the ground ([itex]y=y_\text{ground}[/itex]).

    So, think of Tide's equation as
    [tex]y_\text{ground} = v_{y0}t_\text{ground} - gt_\text{ground}{}^2/2[/tex]
     
  11. Sep 26, 2004 #10
    My way is easier.

    Format, it is moving but at the summit of its trajectory its vertical velocity is zero. Therefore, you can consider only the vertical velocity to find time using
    [tex]v_{f}=v_{i}+at[/tex]
    Solve for t and multiply by two (downward trip as well).
     
  12. Sep 26, 2004 #11
    Sirus is right...but remember that in this formula Vi= initial velocity in y...so it is not the velocity that is given in the problem. Viy = 35m/s sen 40.
    vf would be zero... solve for t and that will be the time to reach the maximum height . Then multiply by two to consider also the time to fall from that maximum height. (in proyectiles the time to go up equals the time to go down)
     
  13. Sep 27, 2004 #12
    Just use the SUVAT equations.

    Consider vertically:
    u = 35sin40, S = 0, a = -9.8, t = ?
    You'll get a quadratic by using the correct formulae. One solution is t=0 (at the start). The other is when the projectile touches the ground again (the time of flight).
     
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