Hi Janet.

Notice that the horizontal aspect of the football's path is entirely irrelevant here - we are just looking at vertical motion. So the way to approach this problem is to use the standard equation for constant acceleration of linear motion. That equation looks like this:

$$s = s_i + v_i\Delta t + \frac 1 2 a(\Delta t)^2$$

In this problem, $s$ represents the vertical distance from the ground to the ball. $s_i$ is the initial distance (given as 2 meters), $v_i$ is the initial vertical speed the ball is thrown upwards with (we don't know that). $\Delta t$ is the time since the ball was thrown, in seconds, and $a$ is the acceleration that gravity provides. Since gravity is driving the ball downwards, it makes sense to use it as a negative value. On earth, this value is -9.8 $\frac m {s^2}$. So let's substitute all that into the equation and simplify it.

$$s = 2 + v_i\Delta t + \frac 1 2 (-9.8)(\Delta t)^2$$

$$s = 2 + v_i\Delta t - 4.9(\Delta t)^2$$

The questions posed give you either the time (that's $\Delta t$) or height (that's $s$) and ask you to find the other. So to find that we need to first determine the only other variable in the equation that we don't yet know: $v_i$. Once you know $v_i$, you can substitute it into the equation above, then plug in for either $s$ or $\Delta t$ and
quickly calculate the answers.

Fortunately, the question starts by giving you enough information to find $v_i$: it tells you two different situations where you know both $s$ and $\Delta t$. So you can create two equations:

"After 1 second it is 6.9 meters high"

$$6.9 = 2 + v_i(1) - 4.9(1)^2$$

"After another second it is 2 meters high"

$$2 = 2 + v_i(2) - 4.9(2)^2$$

Actually, you don't really need to use both equations - you can solve either one to find $v_i$ (I leave that to you). Finding the answer to the two "How high is it after" questions is now easy: you can just plug into the equation what you know and calculate $s$. The question asking "When" is trickier, because you have to solve a quadratic equation. Try using the quadratic formula for this - remember you will get two answers from the quadratic formula usually - you have to figure out if both results make sense for the physical reality of the problem or not!

Hope this helps,
Stephen La Rocque