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Projectile launch point problem

  1. Oct 29, 2006 #1
    I'm having some difficulty with a projectiles question.

    "A projectile is fired with an initial speed of 26.0 m/s at an angle of 40.0 degrees above the horizontal. The object hits the ground 8.00 s later."

    The first part of the question is: "How much higher or lower is the launch point relative to the point where the projectile hits the ground?
    Express a launch point that is lower than the point where the projectile hits the ground as a negative number."

    Which I already found out using y=(viy)(t)-(1/2)(g)(t^2)

    Substituting the given numbers, I get -180 m, which would be the case if the landing point was below the launch point. Since the question wants a launch point lower than the landing point, the answer is 180 m.

    The second part, where I'm having trouble: "To what maximum height above the launch point does the projectile rise?"

    What I have,

    x=(26)(cos(40))(t)
    y=(26)(sin(40))(t)-(1/2)(g)(t^2)

    I can't substitute 180 m or 8 s because those numbers apply to the landing point, and I'm looking for ymax so I end up with three unknowns in two equations, which I cannot solve. I don't know how to set this up so that I can solve it.

    Thanks for any help!
     
    Last edited: Oct 29, 2006
  2. jcsd
  3. Oct 29, 2006 #2

    rsk

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    To find maximum height, remember that the vertical component of velocity at the maximum height is zero.

    Does this help?
     
  4. Oct 29, 2006 #3

    Hootenanny

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    Staff Emeritus
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    Gold Member

    Hint: Try using the formulae;

    [tex]v_{f}^{2}=v_{i}^{2}+2ax[/tex]

    :wink:

    Edit: Beaten to it.
     
  5. Oct 31, 2006 #4
    Whoops! Should post here to let you know I used your hints to figure it out, so thanks!
     
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