Hello, Tiger!

Here's part of it . . .

You see a 10m tall building nearby and you want to throw a snowball onto its roof.

You notice a telepone line running accoss the possible path of your snowball.

The telephone line is 20 m off the ground and is 5 m in front of the building.

You decide to throw the snowball so that is just goes over the telepohne line

and just reaches the roof of the building.

How far from the building should you stand

and with what initial velocity should you throw the ball? Code:

B(0,20)
o
D * | * C
(-5,10)o | o(5,10)
: | :
* : | :
: |20 :10
* : | :
: | :
: | :
A o - - + - - - - - o - - - - - + - - -
(x,0) O 5

The ball is thrown from $\displaystyle A$, passes over the line at $\displaystyle B$, and strikes the roof at $\displaystyle C.$

Place the Origin at the base of the telephone pole.

Then we have points: .$\displaystyle A(x,0),\;B(0,20),\;C(5,10)$

Since the parabolic trajectory is symmetric, the snowball passes through $\displaystyle D(-5,10) $

The general equation for a parabola is: .$\displaystyle f(x) \:=\:ax^2 + bx + c$

Using the three points, we have:

. . $\displaystyle \begin{array}{ccccc}(5,10) & 25a + 5b + c &=& 10 \\ (\text{-}5,10) & 25a - 5b + c &=& 10 \\ (0,20) & 0 + 0 + c &=& 20 \end{array}$

Solve the system of equations: .$\displaystyle a = -\frac{2}{5},\;b = 0,\;c = 20$

. . Hence, the function is: .$\displaystyle f(x) \;=\;-\frac{2}{5}x^2 + 20$

Where is point $\displaystyle A$?

. . (When is the snowball on the ground?)

Set $\displaystyle f(x) = 0\!:\;\;-\frac{2}{5}x^2 + 20\:=\:0 \quad\Rightarrow\quad x^2 \:=\:50 \quad\Rightarrow\quad x \:=\:\pm5\sqrt{2}$

So you should stand about 7 m from the pole, about 12 m from the building.