Results 1 to 6 of 6

Math Help - projectiles

  1. #1
    Member
    Joined
    May 2011
    Posts
    169

    projectiles

    A Rocket R is fired from ground with inital speed s to the horizontal at an angle \alpha. There is a resistive force F=-mkv where m is the mass and v is the vector (v1,v2) and K is a constant. Show the equations of motion are

    d/dt(v1)+kv1=0 and d/dt(v2)+kv2= -g

    My answer. First in the horizontal direction there is no resultant force so I don't get it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Duke View Post
    A Rocket R is fired from ground with inital speed s to the horizontal at an angle \alpha. There is a resistive force F=-mkv where m is the mass and v is the vector (v1,v2) and K is a constant. Show the equations of motion are

    d/dt(v1)+kv1=0 and d/dt(v2)+kv2= -g

    My answer. First in the horizontal direction there is no resultant force so I don't get it

    That is not true. The air resistance is given by

    F=-mkv=-mk(v_1\mathbf{i}+v_2\mathbf{j})

    This force is proportional to the velocity in both of the x and y directions.

    So in the y direction we have by newtons 2nd law

    m\frac{dv_2}{dt}=-mg-kmv_2

    and in the x

    m\frac{dv_1}{dt}=-kmv_1
    Last edited by TheEmptySet; May 26th 2011 at 08:14 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2011
    Posts
    169
    How do you know the sign of kmv? Doesn't it needs to be negative.

    Also I need to state inital conditions for v and r (vectors)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    First I see why you are confused I factored out the minus signs in the force but forgot to put them in the final answer. Yes I had a typo and have edited the above post.

    The initial conditions are implied not explicit given.

    Assumption 1 Since we are free to setup our own coordinate system lets start the rocket at the origin. That would give the initial conditions.

    x(0)=0 \quad y(0)=0

    Also since we are launching the rocket its initial velocity should also be zero

    v_1(0)=0 \quad v_2(0)=0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2011
    Posts
    169
    Thanks. If I define up as positive, do the signs change?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Duke View Post
    Thanks. If I define up as positive, do the signs change?
    No this set of equations defines up as positive in the y direction and right as positive in the x direction. You see this from the fact that -mg has a negative sign. If down were positive this would give the term mg.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Projectiles
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: November 21st 2010, 07:13 AM
  2. Projectiles
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 8th 2010, 02:46 AM
  3. DEs and projectiles
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 20th 2009, 07:14 PM
  4. projectiles?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 16th 2009, 09:15 AM
  5. Projectiles again
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 4th 2009, 08:23 AM

Search Tags


/mathhelpforum @mathhelpforum