Urgent Help Quadratic Fuctions/Relations

**TH1** · July 26th 2007, 10:35 AM

Heres the problem

Your a US general serving in Iraq and have been given orders to bomb a target 20km away. The missile launcher has been programed with an equation showing the parabola trajectory of the missile. If the equation is given in the general form y=ax2+bx+c how can you know if it will hit the target.
Give written instructions for how you would use the equation to answer the question. Give an example of an equation that would hit the target.

For the question i have to write written instructions on how to use the equation to see if it will hit the target.

For my work all I know is that I have to put 20 in the equation
y=ax2+20x+c and I think that ax2 and c are some variables like the wind and speed but I am not sure
Can anyone help me on this I only have less than 8 hours

**Soroban** · July 26th 2007, 11:27 AM

Hello, TH1!

You're a US general serving in Iraq and have been given orders to bomb a target 20km away.
The missile launcher has been programed with an equation
showing the parabolic trajectory of the missile.
If the equation is given in the general form $y \:=\:ax^2+bx+c$ ,
how can you know if it will hit the target?

Code:

        |
        |          ***
        |      *         *
        |   *               *
        |
        |*                     *
        |
    - - * - - - - - - - - - - - * - -
      (0,0)                  (20,0)

We know two points on the parabola: . $(0,0)$ and $(20,0)$ .

Plug them into the general equation: . $y \;=\;ax^2 + bx + c$

$(0,0)\!:\;\;0 \;=\;a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad c \,=\,0$

. . The equation (so far) is: . $y \;=\;ax^2 + bx$

$(20,0)\!:\;\;0 \;=\;a\!\cdot\!20^2 + b\!\cdot\!20\quad\Rightarrow\quad b \:=\:-20a$

. . The equation is: . $\boxed{y \;=\;ax^2 - 20ax}$

Note: Since we want a down-opening parabola, $a < 0$ .

If the general doesn't care how high the missile goes,
. . $a$ can be any negative quantity.

**topsquark** · July 26th 2007, 01:17 PM

Originally Posted by Soroban

Hello, TH1!

Code:

        |
        |          ***
        |      *         *
        |   *               *
        |
        |*                     *
        |
    - - * - - - - - - - - - - - * - -
      (0,0)                  (20,0)

We know two points on the parabola: . $(0,0)$ and $(20,0)$ .

Plug them into the general equation: . $y \;=\;ax^2 + bx + c$

$(0,0)\!:\;\;0 \;=\;a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad c \,=\,0$

. . The equation (so far) is: . $y \;=\;ax^2 + bx$

$(20,0)\!:\;\;0 \;=\;a\!\cdot\!20^2 + b\!\cdot\!20\quad\Rightarrow\quad b \:=\:-20a$

. . The equation is: . $\boxed{y \;=\;ax^2 - 20ax}$

Note: Since we want a down-opening parabola, $a < 0$ .

If the general doesn't care how high the missile goes,
. . $a$ can be any negative quantity.

For the record:
So far we have $y = ax^2 - 20ax$ .

Take the time derivative:
$\frac{dy}{dt} = 2ax \frac{dx}{dt} - 20a \frac{dx}{dt}$

and again:
$\frac{d^2y}{dt^2} = 2a \frac{dx}{dt} + 2ax \frac{d^2x}{dt^2} - 20 a \frac{d^x}{dt^2}$

Now for some Physics. In any projectile motion near the Earth's surface we know that
$\frac{d^2y}{dt^2} = -g$
(I am taking the +y direction to be upward.)
and
$\frac{d^2x}{dt^2} = 0$

Hence:
$-g = 2a \frac{dx}{dt}$

Now, $\frac{dx}{dt} = v_0 \cdot cos(\theta)$ is a constant and is the component of the initial projection velocity in the x direction.

Thus:
$-g = 2a v_0 \cdot cos(\theta)$

Finally:
$a = \frac{-g}{2v_0~cos(\theta)}$

-Dan

**TH1** · July 26th 2007, 01:26 PM

Thank you so much soroban I got a couple questions since the final equation is y=ax2-20ax would it be just left it right at that?I wouldnt have to change anything on that equation and there would be no c or b in that equation?Would that equation actually tell you if the missile hits the target?
Also I have to show my work but the more important thing is that I have to write it in sentences explaining how to use the equation y=ax2+bx+c. So all that math and work I have to put into English so would it be right if i write this?

Two points are already given in the question which are (0,0) (20,0) since its launched from ground and the target is 20km away. Put the 2 points into the equation

the equation now is y=ax2+bx+c

The equation is y=ax2-20ax

Would that work on trying to explain it?

Topsquark I am in grade 10 so my teacher isnt expecting any physics or the type of work you did and would probably suspect something if I used what you did since she never taught us anything like that. Also isnt what Soroban did right also because its far more reasonable for grade 10 and also my teacher taught stuff like that

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