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 $\displaystyle A$.

The target is at $\displaystyle C\!:\;AC = 15$

The sling is 1 meter long: .$\displaystyle AB = 1$

When the projectile is released at $\displaystyle B$, it flies off *tangent* to the circle.

. . Hence: .$\displaystyle \angle ABC = 90^o$

We want the distance $\displaystyle BC.$

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

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