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Homework Help: Solution to quadratic equation doesn't look right

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    distance = rate * time

    3. The attempt at a solution

    So the speed of the ball I have as:

    [t = time]

    5 + 32t

    Which I believe makes sense because the ball is initially traveling @ 5 ft/sec, and after 1 second, the ball is traveling 37 ft/sec, then after 2 seconds the ball is traveling @ 69 ft.sec, etc.

    I assume we use d = rt (which is distance = rate * time).

    So the total distance the ball travels is 85 feet which is equal to the total rate of 5 + 32t * time. The equation I set up like this:

    85 = (5 + 32t) * t

    I set up the quadratic like so:

    32t^2 + 5t - 85 = 0

    I solved it and got:

    (1/64) * (sqrt(10905) - 5) which is approx. 1.55.

    I worked it out to check, and 1.55 seconds is much too low for the value:

    5 feet + (32 feet * 2 seconds) = 69 feet

    So after 2 seconds, the ball has only traveled 69 feet, while the quadratic equation states that the ball has traveled 85 feet after 1.55 seconds.

    What gives?
  2. jcsd
  3. Feb 22, 2012 #2


    User Avatar
    Homework Helper

    You're forgetting that 5+32t represents speed, not distance.
  4. Feb 22, 2012 #3


    Staff: Mentor

    This sounds like a made-up problem that doesn't have much connection to reality. After the ball leaves the pitcher's hand, the only acceleration on it is due to gravity, which acts straight down. Maybe the ball happens to have a propellor or rocket motor attached? Is this the exact wording of the problem?
  5. Feb 23, 2012 #4


    User Avatar
    Science Advisor

    Because this is a two dimensional problem. you should break it into horizontal and vertical components. As Mark44 said there is an acceleration of -9.81 m/s^2 in the vertical direction and no acceleration in the horizontal direction.
  6. Feb 23, 2012 #5

    Char. Limit

    User Avatar
    Gold Member

    Or since he's using feet and seconds, an acceleration of about -32 ft/s^2 vertical and 0 horizontal.
  7. Feb 23, 2012 #6


    Staff: Mentor

    As I said before, this is really a flaky problem, with almost no grounding in reality. I still wonder if the OP is showing the correct statement for this problem.

    Has anyone noticed that the ball's speed is 5 ft/sec? That works out to about 3.4 mi/hr. If I start walking toward the wall at the same time the ball is thrown, I'll get there first (I can walk faster than 3.4 mph).

    Furthermore, I can't imagine any scenario in which the ball would actually get to the wall, inasmuch as it will fall ~4600 ft during its flight.
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