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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?

Respuesta :

Luv2Teach

Answer:

60 meters

Step-by-step explanation:

The standard form for parabolic motion is

[tex]h(x)=-5x^2+v_{0}x+h_{0}[/tex]

where [tex]v_{0}[/tex] is the initial upwards velocity and [tex]h_{0}[/tex] 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.

LucianoBordoli

Answer:

The height of the object at the time of the launch is [tex]60m[/tex]

Step-by-step explanation:

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

[tex]h(x)=-5x^{2}+20x+60[/tex]

For example, after [tex]x=3s[/tex] from the launch the height of the object is :

[tex]h(3s)=-5.(3^{2})+20.(3)+60=75[/tex]

[tex]h(3s)=75m[/tex]

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 [tex]x=0s[/tex] because that is the instant where the object is launched.

If we use [tex]x=0s[/tex] in [tex]h(x)[/tex] ⇒

[tex]h(0s)=-5.(0^{2})+20.(0)+60=60[/tex]

We find that the height of the object at the time of launch is [tex]60m[/tex]

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