Results 1 to 3 of 3

Math Help - Grade 11 High School Circles Question

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    Grade 11 High School Circles Question

    I really do not understand how to do this:

    "An athlete uses a sling to throw a projectile at a target 15m away. He spins the sling in a circle above his head, then releases the projectile. If the sling is 1 m long, at what distance from the target must the athlete release the projectile so that it will hit the target?"

    Answer: ~14.97m

    My only problem is that I have no idea how the answer book came up with that number!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, s3a!@

    It took me a while to catch on . . .


    An athlete uses a sling to throw a projectile at a target 15m away.
    He spins the sling in a circle above his head, then releases the projectile.
    If the sling is 1 m long, at what distance from the target must the athlete
    release the projectile so that it will hit the target?

    Answer: ~14.97m
    Code:
                  * * *    B
              *           *
            *           /   * o
           *           /1    *    o
                      /               o
          *          /        *           o
          *         * - - - - * - - - - - - - o C
          *         A         *  15
    
           *                 *
            *               *
              *           *
                  * * *

    The athlete is at A.
    The target is at C\!:\;AC = 15
    The sling is 1 meter long: . AB = 1

    When the projectile is released at B, it flies off tangent to the circle.
    . . Hence: . \angle ABC = 90^o

    We want the distance BC.

    From Pythagorus: . BC^2 + 1^2 \:=\:15^2 \quad\Rightarrow\quad BC^2 \:=\:15^2 - 1^2 \:=\:225 - 1 \:=\:224

    Therefore: . BC \;=\;\sqrt{224} \;=\;14.96662955 \;\approx\;14.97 m.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597
    Ok thanks for the help! I get a large chunk of it now but I still have minor confusion. What I am confused with is how do you know/assume that we have the hypotenuse and base of the right triangle? Like what I was thinking is 15m-1m or even just 15m. Like can you break down the words a little bit more? Sorry if I am asking for too much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. high school algebra question
    Posted in the Algebra Forum
    Replies: 8
    Last Post: March 10th 2010, 06:59 AM
  2. Replies: 1
    Last Post: November 30th 2008, 11:00 AM
  3. Replies: 16
    Last Post: May 3rd 2008, 01:09 PM
  4. high school
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 29th 2007, 08:49 PM
  5. Replies: 4
    Last Post: November 30th 2007, 09:52 AM

Search Tags


/mathhelpforum @mathhelpforum