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yes

-4.9t^2 + 19.67t +5.06 = 0
Do you see how I got that?

Umm I Think So

Adding Correcct? 1.43 + 4.26 TO BOTH SIDES

-4.9t^2 + 19.67t - 3.46 = 0
There... that's better. :P

That should make more sense now. OK so far?

YES

i Remember

What we can definitely say is that:
a = -4.9
b = 19.67
and c = -3.46
for when we plug in.

\[\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-19.67\pm\sqrt{19.67^2-4(-4.9)(-3.46)}}{2\cdot -4.9}\]

Exactly Correct

Perfect! Now to simplify that horrid looking nightmare. I'll use a calculator now...

One answer is t = 0.184... the other....

t = 3.830

Yes

But we should at least celebrate our progress so far. YAY!!! :)

Yay !!!!!!!!!!!!! Lol

Let's not use the -4.9t^2 + ... mess, instead I'll check the other equation:
-1.43t + 4.26

Right Thats Just Makes Us Go To Far.

OK, plugging in t = 0.184, I see h = 3.99
But t = 3.830, I see h = -1.22

No I Didnt See It , i Just Re-Read The Problem WOW

Almost Missed It

Yes .

Well I Have A Question Mind Helping With Another Onne ?

Go ahead and close this one and post a new one, so it doesn't get too cluttered. See you there! :)

Ok