Results 1 to 3 of 3

Math Help - falling body quadradic function

  1. #1
    Senior Member
    Joined
    Oct 2008
    Posts
    323

    falling body quadradic function

    The general motion equation is given by:

    d(t)=1/2 gt^2+v_ot +d_o where:

    d(t) is the distance traveled, in metres
    g is the acceleration due to gravity in m/s2
    t is the time, in seconds (s)
    vo is the initial velocity of the moving object
    do is the initial height in metres.

    An object is dropped from a height of 1,000 m, on Mars, where the acceleration due to gravity is 3.92 m/s2. Find, to two decimal places, the time it will take for the object to hit the surface of Mars.


    i get 1.96t^2, is it right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by william View Post
    The general motion equation is given by:

    d(t)=1/2 gt^2+v_ot +d_o where:

    d(t) is the distance traveled, in metres
    g is the acceleration due to gravity in m/s2
    t is the time, in seconds (s)
    vo is the initial velocity of the moving object
    do is the initial height in metres.

    An object is dropped from a height of 1,000 m, on Mars, where the acceleration due to gravity is 3.92 m/s2. Find, to two decimal places, the time it will take for the object to hit the surface of Mars.


    i get 1.96t^2, is it right?
    s = 1000m
    u = 0
    a = 3.92
    t = t

    We use s = ut + \frac{1}{2}at^2

    1000 = 0 + \frac{1}{2}3.92t^2

    t = \sqrt{\frac{2000}{3.92}}

    edit: yes, the coefficient of t^2 is 1.96
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, William!

    Their formula isn't accurate . . .
    They should have explained about "upward is positive", etc.


    The general motion equation is given by: . d(t)\:=\:{\color{red}-}\tfrac{1}{2}gt^2+v_ot +d_o where:

    d(t) is the distance traveled, in metres
    g is the acceleration due to gravity in m/sē
    t is the time, in seconds,
    v_o is the initial velocity of the moving object,
    d_o is the initial height in metres.

    An object is dropped from a height of 1,000 m, on Mars,
    where the acceleration due to gravity is 3.92 m/sē.

    Find, to two decimal places, the time it will take for the object to hit the surface of Mars.

    i get 1.96t^2 .
    . . . You have a {\color{blue}t^2} in there?

    We are given: . g = 3.92,\;v_o = 0,\;d_o = 1000

    . . The equation is: . d(t) \;=\;-1.96t^2 + 1000

    "Hit the surface" means d = 0\!:\;-1.96t^2 + 1000 \:=\:0\quad\Rightarrow\quad t^2 \:=\:\frac{1000}{1.96}

    . . Therefore: . t \:=\:\sqrt{\frac{1000}{1.96}} \;\approx\; 22.59\text{ seconds}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. falling body
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 25th 2010, 11:11 PM
  2. Falling body problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 18th 2009, 09:15 AM
  3. Diff EQ Falling body
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2008, 09:05 PM
  4. The falling body
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 16th 2007, 05:05 PM
  5. free body falling
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 8th 2007, 12:02 PM

Search Tags


/mathhelpforum @mathhelpforum