Results 1 to 4 of 4

Math Help - Free Fall Differential Equations Problem - I am stuck

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    2

    Free Fall Differential Equations Problem - I am stuck

    Question:

    An Object of mass 5.00 kg is given an initial downward velocity of 74.5 m/sec and then allowed to fall under the influence of gravity. Assume the force due to air resistance (in N) on this object is twice its speed. If it hits the ground in exactly 10 seconds, how many meters above the ground was the object when it was released?

    The answer should be 368.

    ---
    The part I am stuck in this problem is what will be the drag co-efficient? It says "Assume the force due to air resistance (in N) on this object is twice its speed." - This is the part I DO NOT understand at all!!!

    Please help! Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Setting x(t) the quote of the onbject the equation describing the 'fall' is...

    m\cdot x^{''} = -m\cdot g +2\cdot x^{'} \rightarrow x^{''} -\frac {2}{m}\cdot x^{'}= -g (1)

    The (1) is a linear second order constanr coefficients DE with initial conditions x(0)=x_{0} [ x_{0} is unknown] and x^{'}(0)= -v_{0} and can be solved in standard way. After that the computation of x_{0} is comfortable...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,398
    Thanks
    1327
    Quote Originally Posted by arshadmath View Post
    Question:

    An Object of mass 5.00 kg is given an initial downward velocity of 74.5 m/sec and then allowed to fall under the influence of gravity. Assume the force due to air resistance (in N) on this object is twice its speed. If it hits the ground in exactly 10 seconds, how many meters above the ground was the object when it was released?

    The answer should be 368.

    ---
    The part I am stuck in this problem is what will be the drag co-efficient? It says "Assume the force due to air resistance (in N) on this object is twice its speed." - This is the part I DO NOT understand at all!!!
    Are you serious? "Twice" means "multiply by 2"! The air resistance force is -2v. The total force, then, is -mg- 2v, due to gravity and air resistance.
    m\frac{dv}{dt}= -mg- 2v with v(0)= -74.5. You should be able to solve that equation for v(t). Integrating that will give x(t), the height at time t, with, of course, a constant of integration. Use the fact that x(10)= 0 to find that constant and then find x(0), the initial height.

    Please help! Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2010
    Posts
    2
    lol! But thats what the confusion was; I had no idea that we will call it -2v lol - it now looks so simple after both of yours' help! lol

    Ahh.........I wish that the authors would start writing GOOD books with EXAMPLES in them!! It will be costly lol, but atleast we will understand SOMETHING!!

    lol!! THanks to BOTH of you - a LOT!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. free fall story problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 1st 2011, 10:01 AM
  2. Free fall motion problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 9th 2009, 04:16 AM
  3. Free fall
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 9th 2009, 08:51 PM
  4. Free Fall Lmit/Derivative problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 16th 2007, 08:33 PM
  5. A problem about free fall
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: July 1st 2007, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum