Generally, negative values of time do not have any meaning. Also, we need to assume the projectile hits the ground and then stops - it does not go underground.

So we need to calculate when it is going to hit the ground. This will be when *h* = 0. So we solve:

20

t− 4.9t^{2}= 0

Factoring gives:

(20 − 4.9

t)t= 0

This is true when

`t = 0\ "s"`,

or

`t=20/4.9 = 4.082 text(s)`

Hence, the **domain** of the function *h* is

"all real values of

tsuch that `0 ≤ t ≤ 4.082`"

We can see from the function expression that it is a parabola with its vertex facing up. (This makes sense if you think about throwing a ball upwards. It goes up to a certain height and then falls back down.)

What is the maximum value of *h*? We use the formula for maximum (or minimum) of a quadratic function.

The value of *t* that gives the maximum is

`t = -b/(2a) = -20/(2 xx (-4.9)) = 2.041 s `

So the maximum value is

20(2.041) − 4.9(2.041)

^{2}= 20.408 m

By observing the function of *h*, we see that as *t* increases, *h* first increases to a maximum
of 20.408 m, then *h* decreases again to zero, as expected.

Hence, the **range** of *h* is

"all real numbers, `0 ≤ h ≤ 20.408`"

Here is the graph of the function *h*:

Domain: `0<=x<=4.08`

Range:

`0<=h<=20.4

`0<=h<=20.4

Get the Daily Math Tweet!

IntMath on Twitter