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

Answer:

60 meters

Step-by-step 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

Answer:

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

Step-by-step 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 $60m$