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9 + 10?

19

STOP IT, GOD DAMN IT !

`PART 1:`
1) What is the initial speed of the cannonball?
\(\sqrt{33^2(m/s)+25^2(m/s)~}=41(m/s)\)

`PART 3`
3) What is the maximum height the cannonball goes above the ground?

I just want to do it a plain way using some familiar formula

oh, it should be 25 not 33

because we are computing vertical displacement we must use vertical velocity

Okay, I am getting through it with a painful struggle-:(:)

good, nothing to feel bad about
these things take time

Like nothing else in this world.... :)
But, true that

Well, the fact that my teacher is awful gives me more credit

4) How far from where it was shot will the cannonball land?

you should not get multiple "t" values
the vertex is a "single" point
|dw:1441337672713:dw|

31.855249745=25t−(9.81/2)t^2

use the exact value for \(x\), then u will get a single \(t\ value

use the exact value for \(x\), then u will get a single \(t\) value

well, I got a single approximation for t (as for now because I increased my value-precision)

2.54 seconds

looks fine

oh wait 2.54 is half of the whole time, right?

BINGO

Yes, 2.54 = time at which the particle reaches max height

Yes, so from max back to v=0 another 2.54s

Thank you!

oh, I also tried t=3.7 without subtracting and it was still wrong