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:
20t − 4.9t2 = 0
(20 − 4.9t)t = 0
This is true when
`t = 0\ "s"`,
`t=20/4.9 = 4.082 text(s)`
Hence, the domain of the function h is
"all real values of t such 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: