# A rocket fired into the air is modeled by the function h(t)=-16t^2 + 160t. How many seconds is the rocket in the air?

Jun 8, 2016

The rocket will be in the air for $10$ seconds.

#### Explanation:

We have to find out the positive values of $x$ for which $h \left(x\right) \ge 0$.

We can do it algebraicly:

$- 16 {t}^{2} + 160 t \ge 0$

$- 16 t \cdot \left(t - 10\right) \ge 0$

${t}_{1} = 0 , {t}_{2} = 10$

Since the coefficient next to ${x}^{2}$ is nagative ($- 16$), the solution of this inequality is t in <0;10>, so the rocket's flight will last $10$ seconds.

We can also solve this task using graph of the function:

graph{-16x^2+160x [-50, 50,-20, 428]}

From this graph we clearly see, that the height is not lower than zero for t in <0;10>

Jun 8, 2016

$10 s$

#### Explanation:

Given is
$h \left(t\right) = - 16 {t}^{2} + 160 t$
It has not been explicitly given but it is presumed that $h$ is the height of rocket and $t$ is time in seconds.

Clearly the rocket will be in air between the time interval from ${t}_{0} = 0$ and $t = {t}_{1}$ when height of the rocket is zero.
Setting the height $h \left(t\right) = 0$ and solving for $t$
$- 16 {t}^{2} + 160 t = 0$
Factorizing we obtain
$- 16 t \left(t - 10\right) = 0$
We have two values of $t$

1. from first factor
$t = 0$
2. from second factor
$\left(t - 10\right) = 0$
$\implies t = 10$

We obtain ${t}_{1} - {t}_{0} = 10$ seconds is time for which the rocket is in the air.
Graphically
graph{y=-16x^2+160x [-5, 15,-15, 428]}
height $h$ is plotted as $y$ and time $t$ as $x$
We obtain time of flight as $t = 10$