   SEARCH HOME Math Central Quandaries & Queries  Question from Janet, a student: a football is thrown into the air from 2 meters high. After 1 second it is 6.9 meters high. After another second it is 2 meters high. How high is it after 1.02 seconds and .05 seconds? when will it reach 5.136 meters? I am not sure how to extract the numbers I need and what equations to use them in? 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     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.