# You toss a ball into the air from a height of 5 feet velocity of the ball is 30 feet per second. You catch the ball 6 feet from the ground. How do you use the model 6=-16t^2+30t+5 to find how long the ball was in the air?

##### 1 Answer
Nov 22, 2017

$t \approx 1.84$ seconds

#### Explanation:

We are asked to find the total time $t$ the ball was in the air. We are thus essentially solving for $t$ in the equation $6 = - 16 {t}^{2} + 30 t + 5$.

To solve for $t$ we rewrite the equation above by setting it equal to zero because 0 represents the height. Zero height implies the ball is on the ground. We can do this by subtracting $6$ from both sides

$6 \cancel{\textcolor{red}{- 6}} = - 16 {t}^{2} + 30 t + 5 \textcolor{red}{- 6}$

$0 = - 16 {t}^{2} + 30 t - 1$

To solve for $t$ we must use the quadratic formula:
$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where $a = - 16 , b = 30 , c = - 1$

So...

$t = \frac{- \left(30\right) \setminus \pm \sqrt{{\left(30\right)}^{2} - 4 \left(- 16\right) \left(- 1\right)}}{2 \left(- 16\right)}$

$t = \frac{- 30 \setminus \pm \sqrt{836}}{- 32}$

This yields $t \approx 0.034 , t \approx 1.84$

Notice: What we ultimately found were the roots of the equation
and if we were to graph the function $y = - 16 {t}^{2} + 30 t - 1$ what we will get is the path of the ball.

https://www.desmos.com/calculator/vlriwas8gt

Notice in the graph (see link), the ball is shown to have touch the ground twice at the two $t$ values we initially found but in the problem we throw the ball from an initial height of $5 \text{ft}$ so we can disregard $t \approx 0.034$ because that value is implying that the ball was thrown at an initial height of zero which it wasn't

Thus, we are left with $t \approx 0.034$ which is the other root which on the graph, represent the time for the ball to hit ground giving us the total time of flight (in seconds I presume).