Quadratics/parabolas

[sand stag]

How would I find the answer? Does 64 = c? 😭

[gentle hawk]

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[sand stag]

The context: “for a science project you and your friends launch a rocket from ground level with an initial vertical velocity of 80 feet per second” .

So the equation is: h(t) = -16t^2 + 80t

$ h(t) = (-16t^(2)) + (80t)$

$h(t) = (-16t^2) + (80t)$

[brittle pike]

𝕃𝕦𝕤𝕥𝕣𝕠𝕦𝕤

[nimble kiln]

Here's what I did

[sand stag]

Wait a minute…could I just do -16t^2 +80t - 64?

[nimble kiln]

Set the equation equal to 64

[nimble kiln]

Right.

[sand stag]

Okay

[nimble kiln]

Divide it al by 16 to simplify

And then factor

[sand stag]

Thank you 🙏

[nimble kiln]

meant

-16

What do you get after factoring?

[sand stag]

t = 1

[nimble kiln]

As in

Solutions for t

when y = 64

or in this case h

[sand stag]

t = 4 or t = 1 right?

[nimble kiln]

Correct

So since the t axis stands for the amount of time

And the amount of time would be the distance between those solutions

Subtract 4 from 1 to get the amount of time it was in the air for

[sand stag]

Ohhh okay

[nimble kiln]

Do you understand that?

[sand stag]

Yeah…

[nimble kiln]

Nice

@sand stag Here's a somewhat similar question you could solve

A diver jumps off a platform, and their height in feet at any time would be t in seconds. This is given by the equation: h(t) = -16t^2 +48t + 10

At what times' does the diver reach 30ft in dpeth?

fuck

depth*

[sand stag]

Let me do the math first

[nimble kiln]

Yeah

You can do the math

[sand stag]

I got 3

@nimble kiln

No wait

It’s 1/2

[nimble kiln]

Yeah I got

1/2 and

2 and 1/2

[sand stag]

Wait a minute…I got 5/2

[nimble kiln]

Same thing

[iron nest]

eq looks to be $-16t^2+80t\geq{64}$

[brittle pike]

;( | 追放された興奮

[iron nest]

this is because we are looking at how long this problem is persisting for and thus need to find the time in which the function is greater than or equal to 64

dividing both sides by 16 we get $-t^2+5t\geq{4}$