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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?

User Idoshamun
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2 Answers

4 votes

Answer:

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.

User Dardan
by
6.5k points
4 votes

Answer:

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

User Hamdi
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7.4k points