Results 1 to 5 of 5

Math Help - DEs and projectiles

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    37

    DEs and projectiles

    This problem relates to projectiles without air resistance, but using DEs Solve the IVP to find an expression for v(t) in terms of g and t(no m dependence)

    1st: Given mv'(t)=-mg ; v(0)=v0

    2nd: v'(t)=-g (dividing by m, since there is no mass dependence)
    3rd: integrate v(t)=-9.8t

    I'm stuck here, because I am to use the result from v(t) and solve for the IVP: x'(t)=v(t) ; x(0)=0 (height above ground of the projectile) in terms of v0,g, and t.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by latavee View Post
    This problem relates to projectiles without air resistance, but using DEs Solve the IVP to find an expression for v(t) in terms of g and t(no m dependence)

    1st: Given mv'(t)=-mg ; v(0)=v0

    2nd: v'(t)=-g (dividing by m, since there is no mass dependence)
    3rd: integrate v(t)=-9.8t

    I'm stuck here, because I am to use the result from v(t) and solve for the IVP: x'(t)=v(t) ; x(0)=0 (height above ground of the projectile) in terms of v0,g, and t.
    Could you post the original problem, please?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    37

    problem detail

    A projectile is shut up vertically without taking air resistance into account, considering a basic vertical throw.After being shot, the only force acting upon the projectile is gravity -mg, with negative sign indicating gravity act downwards. Thus Newton's Law gives ma(t)=-mg. With a(t)=v'(t) and use v(0)=vo(not) as initial condition this gives the first order initial value problem for velocity:

    mv'(t)=-mg, v(0)=vo....my original post goes from here.....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by latavee View Post
    A projectile is shut up vertically without taking air resistance into account, considering a basic vertical throw.After being shot, the only force acting upon the projectile is gravity -mg, with negative sign indicating gravity act downwards. Thus Newton's Law gives ma(t)=-mg. With a(t)=v'(t) and use v(0)=vo(not) as initial condition this gives the first order initial value problem for velocity:

    mv'(t)=-mg, v(0)=vo....my original post goes from here.....
    First integrate both sides to get v(t)=-gt+c.

    Applying the initial condition v(0)=v_0, we have v(0)=v_0=-g(0)+c\implies c=v_0.

    Thus, v(t)=-gt+v_0.

    Since x^{\prime}(t)=v(t), it follows that if we integrate both sides of the equation, we have

    x(t)=-\tfrac{1}{2}gt^2+v_0t+c.

    Now, supposing that x(0)=x_0, we have x(0)=x_0=-\tfrac{1}{2}g(0)^2+v_0(0)+c\implies c=x_0.

    Thus, x(t)=x_0+v_0t-\tfrac{1}{2}gt^2.

    In your first post, I saw that you let x_0=0. So it turns out that the equation you're interested in is x(t)=v_0t-\tfrac{1}{2}gt^2.

    (If you're familiar with physics, we take note that this is the kinematics equation s_f=s_0+v_0t+\tfrac{1}{2}at^2, where we have s_0=0 and a=-g)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2009
    Posts
    37

    Thanks

    Ah! Thank you...you made that so understandable!
    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. projectiles?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 16th 2009, 09:15 AM
  4. Projectiles again
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 4th 2009, 08:23 AM
  5. Projectiles
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 4th 2009, 05:24 AM

Search Tags


/mathhelpforum @mathhelpforum