# The function h(t)=-5t^2+20t+2 gives the approximate height, h meters, of a thrown football as a function of the time, t seconds, since it was thrown. The ball hit the ground before a receiver could get near it. a) How long was the ball in the air?

Mar 7, 2015

About $4.1$ seconds.

How long was the ball in the air?
We are told that $t$ represents time in seconds since the ball was thrown, so it started to be 'in the air' at $t = 0$

To answer the question, then, we need to know the time when it stopped being in the air.

We are told that the ball hit the ground. So that's what happened when it stopped being airborne. We need to relate that event to the mathematics we're working with.

What can we say about $h$, the height of the ball when the ball hits the ground? Answer: The height will be $0$ when the ball stops being in the air.

Now translate this back to the mathematics:

The ball is in the air from $t = 0$ until the time $t$ when $h = 0$.
Find the time $t$ that makes $h = 0$.

That means: solve: −5t^2+20t+2=0
We can solve this by solving: $5 {t}^{2} - 20 t - 2 = 0$
(Either multiply both sides of the equation by $- 1$, or add $5 {x}^{2}$ $- 20 x$ and $- 2$ to both sides and then re-write it the other way around)

That's a quadratic equation, so try to factor first.
But don't spend too much time trying to factor, because not every quadratic is easily factorable and that's OK, because we still have the quadratic formula if we need it. We do need it.

$t = \frac{- \left(- 20\right) \pm \sqrt{{\left(- 20\right)}^{2} - 4 \left(5\right) \left(- 2\right)}}{2 \left(5\right)} = \frac{20 \pm \sqrt{440}}{10}$

$= \frac{20 \pm \sqrt{4 \left(110\right)}}{10} = \frac{20 \pm 2 \sqrt{110}}{10}$

$= \frac{2 \left(10 \pm \sqrt{110}\right)}{2 \left(5\right)} = \frac{10 \pm \sqrt{110}}{5}$

We can see that $10 < \sqrt{110} < 11$. In fact ${\left(10 + \frac{1}{2}\right)}^{2} = {10}^{2} + 10 + \frac{1}{4} = 110.25$

Using $10.25$ as an approximation for $\sqrt{110}$, we get :
for the solution $t = \frac{10 - \sqrt{110}}{5}$ we'll get a negative $t$. That doesn't make sense.
The other solution gives
$t \approx \frac{10 + 10.25}{5} = \frac{20.5}{5} = 4.1$ seconds.

So the ball was in the air from $t = 0$ until about $t = 4.1$. The elapsed time is the difference, $4.1$ seconds.