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Homework Help: Rocket travels along parabolic path (RECTANGULAR COMPONENTS)

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    When a rocket reaches an altitude of 40m it begins to travel along the parabolic path (y-40)^2=160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy=180m\s, determine the magnitudes of the rockets velocity and acceleration when it reaches an altitude of 80m.

    2. Relevant equations

    position: r=x*i+y*j (where i and j are the unitvector)
    velocity: v=dr\dt
    acc: a=dv\dt

    3. The attempt at a solution

    lets say i tried for about 4 hours to get a concept to solve this. :(

    first i put in in the equation for the motion path for y=80 to get x=10

    second: position vector for point A (x=0,y=40) and point B (x=10 y=80)

    third: to get the distance betwee A and B i subtract the 2 position vectors.

    fourth: v=dr\dt to get the velocity

    mazbe its the wrong way because the answer should be v=201 m\s and mines was 254 m\s.

    i dont know where my mistake is...is the way of thinking right?
  2. jcsd
  3. Feb 3, 2010 #2
    I don't how you found out dr/dt without separating out the equations for motion on x,y-axes w.r.t time from the trajectory equation.
    Velocity on y axis is constant so y=180t.
    Put in trajectory eqn. to eliminate y then differentiate w.r.t. x. This will be velocity on x-axis. Add vectorially for complete velocity.
    Or you could directly differentiate the trajectory equation w.r.t time. dy/dt is known, dx/dt can be found out and then added vectorially.
  4. Feb 3, 2010 #3


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    Notice that eqn#2 has a variable in it that is defined in eqn#1. If you plug this definition from #1 into #2, what equation do you get for the velocity?
  5. Feb 3, 2010 #4
    ok i tried out this:

    d/dt (y-40)²=d/dt 160x

    => dx/dt=405t-90



    but how can i find the magnitude? since in the equation for v there is still the variable "t".
  6. Feb 3, 2010 #5


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    Can you now find out how much time it takes for the projectile to reach a height of 80m?
    Last edited: Feb 3, 2010
  7. Feb 3, 2010 #6
    may answer would be:

    for the equation of path i put in y=180t and x=10 (when y=80 x=10)

    then i get a quadratic function in t.

    the result for t= 0.4s

    i put that in v and get v=72i+180j

    for magnitude: v=193,7 m/s

    but as result there should be 201 m/s
  8. Feb 4, 2010 #7
    t=(distance on y-axis)/(Speed on y-axis)
    This is 40/180=0.22s.
  9. Feb 4, 2010 #8
    thank you very much!!
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