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!

Projectile in Motion

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data
    A rocket powered hockey puck on frictionless ice slides along the y-axis. The front of the rocket is tilted at an angle from the x-axis. The rocket motor ignities as the puck crosses the origin, exerting a Force(thrust) on the puck.
    A. Find algebraic equation y(x) for the pucks trajectory
    B. Thrust= 2N.
    Mass of rocket/puck togther =1kg
    Intial speed is 2.0m/s^2 along the y-axis. Make graph of f(x) at 45 deg. and -45 deg. fromx=0 to 20m

    2. Relevant equations
    y=y(nought)+vt+.5at^2
    x=x(nought)+vt+.5at^2
    a=F(net)/m

    3. The attempt at a solution
    A. I get: y(x)=V*(sqrt(2x/a*cos))+xtan
    But it is incorrect.
    B. at -45 deg, y(8.5) is suppose to equal 0 but my equation doesnt follow this.
     
    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 5, 2007 #2
    Anyone?...
     
  4. Feb 5, 2007 #3
    Its always difficult to answer these questions without the work shown. What would help is to show the x and y accelerations as a function of angle. In one case the puck will try to fly opposed by gravity, in the -45, the rocket thrust will add to gravity. In both cases the x acceleration is some fraction of the rocket force.
     
  5. Feb 6, 2007 #4
    actually, the puck does not leave the ground; the puck is just in a x/y plane on ice taking on a projectile motion.
    For X: x= .5a(x)cost*t^2, a(x) is acceleration in the x direction.
    Y: y=vt+.5a(y)sin*t^2, a(y) is the acceleration in the y direction.
    I solved for t in my x equation and got t=sqrt(x/a*cos)
    Substituting t into Y I get y(x)=V*(sqrt(2x/a*cos))+xtan.
    a=2m/s^2 so the equation can simplify to
    y(x)=V*(sqrt(x/cos))+xtan.
     
    Last edited: Feb 6, 2007
  6. Feb 6, 2007 #5
    lets call a(x)=a1. So x=.5*a1*cos(theta)*t^2 as initial Vx=0;

    then t=sqrt(2x/(a1*cos(theta))) as in somewhere the .5 got dropped?
     
  7. Feb 6, 2007 #6
    ok, my prof. gave us the solution to this problem today. I did get the generic equation right but still the wrong value. He said:
    y=sqrt((2*1kg*x)\(2N*cos45))+xtan45; this simplyfies to :
    y=2.38*sqrt(x)+x, but how? where does this 2.38 come from? xtan45=1x but what happens to the other terms?
     
  8. Feb 6, 2007 #7
    Scratch that last note, I got it
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Projectile in Motion
Loading...