Question

If an object is thrown straight up into the air with a velocity 144 feet/second, then its height h(t) above theground t seconds later is given by the formulah(t): = - 16t² + 144t-Find the height after 2 and after 5 seconds. [Find h(2) and h(5).]h(2) =h(5) =

Answer 1

The following function gives the height of an object thrown straight up after t seconds:

`h(t)=-16t^2+144t`

It's required to find h(2) and h(5).

a) To find h(2), substitute the value t = 2 into the equation as follows:

`h(2)=-16(2)^2+144(2)`

Operating:

`\begin{gathered} h(2)=-16(4)+288 \\ \\ h(2)=-64+288 \\ \\ h(2)=224 \end{gathered}`

b) Now find h(5):

`h(5)=-16(5)^2+144(5)`

Calculating:

`\begin{gathered} h(5)=-16(25)+720 \\ \\ h(5)=-400+720 \\ \\ h(5)=320 \end{gathered}`

The required heights are:

h(2) = 224

h(5) = 320