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Projectile Motion Homework Help

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the initial velocity and the maximum height

    θ = 28˚
    horizontal range/distance/s = 68 m
    time = 6.30 s

    2. Relevant equations

    suvat equations not so sure which one though

    3. The attempt at a solution

    I think to find the initial velocity I use the equation:
    s=ut+(1/2)at^2

    I think acceleration horizontally is 0 so:

    u(h) = s(h)/t
     
  2. jcsd
  3. Oct 6, 2011 #2
    There is no acceleration in the horizontal direction so you can simply use s=vt.

    Now you can solve for the horizontal component of velocity and you have an angle so using trig you should now be able to solve for the hypotenuse of the initial velocity.

    For the maximum height just remember to only focus on the vertical components and use your kinematics equations.
     
  4. Oct 6, 2011 #3

    SammyS

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    Use the last equation, u(h) = s(h)/t to find the horizontal component of the velocity. Since the horizontal component of the velocity is constant, what does this tell you about the horizontal component of the initial velocity?
     
  5. Oct 6, 2011 #4
    It is the same as the u(h)?
     
  6. Oct 6, 2011 #5
    Lol I don't know any of this...
     
  7. Oct 6, 2011 #6

    SammyS

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    Yes. The horizontal component of the velocity doesn't change, so that is the same as the horizontal component of the initial velocity.

    From that use some trig to find the initial velocity and the vertical component of the initial velocity.
     
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