# Grade 11 High School Circles Question

• November 29th 2008, 07:34 PM
s3a
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?"

My only problem is that I have no idea how the answer book came up with that number!
• November 29th 2008, 08:54 PM
Soroban
Hello, s3a!@

It took me a while to catch on . . .

Quote:

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?

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.

• November 30th 2008, 09:35 AM
s3a
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.