Results 1 to 3 of 3

Math Help - Determining Rates of Change

  1. #1
    !!!
    !!! is offline
    Junior Member
    Joined
    Feb 2008
    Posts
    28

    Determining Rates of Change

    I need help on how to understand and solve this problem:

    A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
    When does it reach its maximum height?
    When does it hit the ground?
    (Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per second^2.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by !!! View Post
    I need help on how to understand and solve this problem:

    A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
    When does it reach its maximum height?
    When does it hit the ground?
    (Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per second^2.
    Take upwards as positive:

    Data:
    u = 48 ft/sec
    v = 0 ft/sec
    a = -32 ft/sec^2
    t = ?

    Substitute the data into v = u + a t and solve for t.


    Data:
    u = 48 ft/sec
    a = -32ft/sec
    x = -432 ft
    t = ?

    Substitute the data into x = ut + 1/2 a t^2 and solve for t.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by !!! View Post
    I need help on how to understand and solve this problem:

    A ball is thrown upward at a speed of 48 feet per second from the edge of a cliff 432 feet above ground. Find its height above ground t seconds later.
    When does it reach its maximum height?
    When does it hit the ground?
    (Note: Give the upward direction as positive, physics tells us that the acceleration of the ball is -32 feet per second^2.
    With an accelerated movement you should know that

    v(t) = a\cdot t ....... If the acceleration is a then the speed after t seconds is v(t).

    d(t) = \frac12 \cdot a \cdot t^2 ....... If the acceleration is a then the distance after t seconds is d(t).

    With your problem:

    v(0) = 48\ \left[\frac{ft}{s}\right]

    v(t) = 48 - 32 \cdot t \ \left[\frac{ft}{s}\right]

    d(t) = 48\cdot t - \frac12 \cdot 32 \cdot t^2 \ [ft] = 48t - 16t^2\ [ft] ....... And therefore the height over ground is calculated by:

    h(t) = d(t) + 432\ [ft] = -16t^2 +48t+ 432\ [ft]

    The graph of d(t) is a parabola opening down. So the highest point is reached at the vertex of the parabola which occurs after t = 1.5 s:

    h(1.5) = 468\ [ft]

    The ball hits the ground if the height is zero:

    -16t^2 +48t+ 432 = 0 ....... Solve for t. You'll get 2 solutions but one of them isn't very plausible. I've got 6.91 s.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rates of Change
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 23rd 2010, 03:15 PM
  2. Rates of change
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 25th 2009, 10:44 AM
  3. Rates of Change
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 2nd 2009, 07:34 AM
  4. Determining Rates
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 19th 2009, 06:05 PM
  5. Rates Of Change
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 22nd 2008, 03:40 AM

Search Tags


/mathhelpforum @mathhelpforum