Question

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?

Answer 1

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

Explanation:

We have to find out the positive values of `x` for which `h(x)>=0`.

We can do it algebraicly:

`-16t^2+160t>=0`

`-16t*(t-10)>=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>`

Answer 2

`10s`

Explanation:

Given is
`h(t)=-16t^2+160t`
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(t)=0` and solving for `t`
`-16t^2+160t=0`
Factorizing we obtain
`-16t(t-10)=0`
We have two values of `t`

1. from first factor
   `t=0`
2. from second factor
   `(t-10)=0`
   `=> 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`