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Mathematics 16 Online
OpenStudy (anonymous):

An object is fired upward from the top of a 200 ft. tower at a velocity of 80 ft/sec. The height of the object, h(t), is modeled by the function h(t) = -16t2 + 80t + 200, where t represent the time in seconds.

OpenStudy (anonymous):

Howdy my friend

OpenStudy (anonymous):

hi :D

OpenStudy (anonymous):

What do you need help with?

OpenStudy (anonymous):

An object is fired upward from the top of a 200 ft. tower at a velocity of 80 ft/sec. The height of the object, h(t), is modeled by the function h(t) = -16t2 + 80t + 200, where t represent the time in seconds. this question

OpenStudy (anonymous):

Right. But there would be more questions concerning that question you just typed.

OpenStudy (anonymous):

You know the equation and everything that's great

OpenStudy (anonymous):

Which time is specified? I bet you are left to use the formula to calculate the height of the object at certain time following the launch

OpenStudy (anonymous):

I have to find the Maximum

OpenStudy (anonymous):

Let's graph.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Find the vertex.

OpenStudy (anonymous):

That is the point at which the parabola reflects.

OpenStudy (anonymous):

Do you have an access to online calculator or would you like me to do it?

OpenStudy (anonymous):

I would like you to show me

OpenStudy (anonymous):

I am on it just a sec

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

y=-16t ^ 21+80t+200

OpenStudy (anonymous):

Type that into desmos calculator. You should get a vertex of (2.5,300)

OpenStudy (anonymous):

It's saying that after 2.5 seconds from the launch the height of the object reaches 300m

OpenStudy (anonymous):

Did you get it or not?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so is 300m

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