Question

On Mars, gravity is less than it is on Earth. If you were to kick a Mars rock at an initial velocity of 38 feet per second, its height h would be modeled by the equation f (t) = -1.9t^2+ 38t where t is time in seconds. How long would the rock be in the air?

Answer 1

To find the time that the rock is in the air, we need to find the times when f(t)=0:

`0=-1.9t^2+38t`

We factor this equation as follows:

`0=t(-1.9t+38)`

And now we apply zero product property, if ab=0, then a=0 or b=0.

In this case:

`\begin{gathered} t=0 \\ or \\ -1.9t+38=0 \end{gathered}`

Since t=0 is the stating time, we only take the second equation, and solve for the time:

`\begin{gathered} -1.9t+38=0 \\ -1.9t=-38 \\ t=(-38)/(-1.9) \\ t=20 \end{gathered}`

Answer: 20 seconds