1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    v=vx*i+vy*j

    v=(405t-90)*i+180j

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

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rocket travels along parabolic path (RECTANGULAR COMPONENTS)
Loading...