Question

An object is launched from a platform. It's height (in meters), x seconds after the launch, is modeled by: h(x) = -5x^2 + 20x + 60. What is the height of the object at the time of launch?

Answer 1

60 meters

Explanation:

The standard form for parabolic motion is

`h(x)=-5x^2+v_(0)x+h_(0)`

where
`v_(0)` is the initial upwards velocity and
`h_(0)` is the initial launching height. If I am understanding your question, this is what you are looking for. So the height AT the time of launch was 60 meters.

Answer 2

The height of the object at the time of the launch is
`60m`

Explanation:

We know that the height in meters, x seconds after the launch is modeled by the following function :

`h(x)=-5x^(2)+20x+60`

For example, after
`x=3s` from the launch the height of the object is :

`h(3s)=-5.(3^(2))+20.(3)+60=75`

`h(3s)=75m`

If we want to know the height of the object at the time of the launch we will need to find the height of the object at
`x=0s` because that is the instant where the object is launched.

If we use
`x=0s` in
`h(x)` ⇒

`h(0s)=-5.(0^(2))+20.(0)+60=60`

We find that the height of the object at the time of launch is
`60m`