how to plug height and time into a quadratic equation $h=at^2+bt+c$

Question

How to plug height and time into a quadratic equation $$h=at^2+bt+c$$ to model the height of the object $t$ seconds after it is shot into the air? The height is $163.6$ feet the time is $1.2$ seconds.

Is it like this:

$$h=163.6(1.2)^2+163.3(1.2)+0$$

Answer 1

It looks you are actually being asked to find the values of the parameters $a$, $b$ and $c$. You know that:

• at $t=1.2$, $h=163.6$
• at $t=0$, $h=0$ (this is presuming the height is measured relative to where the object was shot from)

You can use the second piece of information to work out $c$. However you have not given enough information to find $a$ and $b$. (You need one more data point to determine $a$ and $b$ uniquely.)

Answer 2

In your equation, $h$ is probably the height, $a$ is twice the acceleration ($-2g$), $b$ is the initial velocity, and $c$ is the initial height. The typical equation used for this is called a kinematic equation

$x = x_0 + v_0 t + \frac{1}{2}a_0t^2$,

where $a_0 = g = 9.8m/s^2$.

On the other hand, if you are just given this equation with no knowledge of the physical situation, you will have to plug in 163.6 for $h$ and 1.2 for $t$ and solve for $a$, $b$, and $c$. Also, you probably know that at $t=0$, $h=0$, so you can use this to find $c = 0$. However, you can't find $a$ or $b$ without more information about the trajectory (the velocity and acceleration at $t=0$, for instance).

Edit:

"After 1.2 seconds the object is 163.6 feet off the ground, at 4 seconds it is 307.8 feet off the ground, and at 5 seconds it is 298.5 feet off the ground"

Now you have three equations:

$163.6 = a(1.2)^2+b(1.2)+c$

$307.8 = a(4)^2+b(4)+c$

$298.5 = a(5)^2+b(5)+c$

By solving this system of equations, you can find the values of a, b, and c