Question

Ali kicks a soccer ball in the air. The pathway of the ball can be modeled by `(t) = -4.9t^2 + 25.1t +1`, where height, `h` is measured in meters and time, `t`, can be measured in seconds. How long will the ball be in the air?

Answer 1

As `t` is time of flight duration then the ball is kicked at `t=0`
At `t=0` the ball is 1 metre above ground level. Thus the ball was not kicked from ground level.
`color(red)("Time of flight from "t=0" is approximately 5.16 seconds")`

Explanation:

This is a quadratic in `t` rather than in `x`

`color(brown)("There is a trap in this question")`

Set `t_0 =0` as the time instant when the ball is kicked.

Consider the standardised form of `y=ax^2 +bx+c`
The thing to note is that the constant `c` is y-intercept. As `c=1` it means that the ball is kicked from 1 metre above ground. Thus the first x-intercept is not at `t_0` but before it. Thus we need to discount the time of the first x-intercept but consider time of flight as starting at `t_0->x=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set `-4.9t^2+25.1t+1=0`

To get rid of the decimal multiply both sides by 10:

`10xx0=-49t^2+251t+10`

`0=-49t^2+251t+10`

Using the formula `t=(-b+-sqrt(b^2+4ac))/(2a)`

`t=(-251+-sqrt(251^2-4(-49)(10)))/(2(-49)`

`t=(-251+-sqrt(64961))/(-98)`

`t=+2.56122...+-2.6007...`

`t~~-0.04 larr" discarded value as before "t_0`

`t~~color(white)(-)5.16 larr" accepted time of flight"`