Math-Based-Physics question.

- idku

Math-Based-Physics question.

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- yolomcswagginsggg

9 + 10?

- anonymous

19

- yolomcswagginsggg

21.

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## More answers

- idku

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.
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- idku

STOP IT, GOD DAMN IT !

- yolomcswagginsggg

xD

- idku

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

- idku

`PART 2:`
2) What is the initial angle θ of the cannonball with respect to the ground?`PART 1:`
\(\theta=\tan^{-1}\left(25/33\right)\approx37.15\)

- idku

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

- idku

\(\Large v^2=u^2+2ax\)
(NOTE: x is vertical displacement)
Ok, so I know that v=0 when the ball reaches maximum height.
My initial vertical velocity is 33m/s.
Acceleration is -9.81a
\(\Large 0^2=33^2+2(-9.81)x\)
\(\Large 33^2=2(9.81)x\)
\(\Large x=55.5\)
and that is wrong...

- idku

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

- idku

oh, it should be 25 not 33

- idku

because we are computing vertical displacement we must use vertical velocity

- idku

\(\Large v^2=u^2+2ax\)
(NOTE: x is vertical displacement)
Ok, so I know that v=0 when the ball reaches maximum height.
My initial vertical velocity is 25m/s.
Acceleration is -9.81a
\(\Large 0^2=25^2+2(-9.81)x\)
\(\Large 25^2=2(9.81)x\)
\(\Large x=31.8\)
and that is CORRECT!

- idku

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

- ganeshie8

good, nothing to feel bad about
these things take time

- idku

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

- idku

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

- idku

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

- idku

\(\Large x = ut + \dfrac{1}{2}at^2\)
\(\Large 31.8 = 25t - \dfrac{9.81}{2}t^2\)
t = 2.44229
t = 2.65455
then I can use:
Horizontal displacement = time × initial horizontal velocity
my problem now is which time to take

- ganeshie8

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

- idku

31.855249745=25t−(9.81/2)t^2

- idku

then I get about same thing
2.54 seconds
http://www.wolframalpha.com/input/?i=31.855249745%3D25t%E2%88%92%289.81%2F2%29t%5E2

- ganeshie8

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

- ganeshie8

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

- idku

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

- idku

2.54 seconds

- ganeshie8

looks fine

- idku

(Horizontal displacement) = (time) × (initial horizontal velocity)
(Horizontal displacement) = (2.54s) × (33m/s)=83.82s
but the practice thingy tells me that is wrong

- idku

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

- idku

Yup.
(Horizontal displacement) = (time) × (initial horizontal velocity)
(Horizontal displacement) = \(\color{red}{\rm 2 \times }\) (2.54s) × (33m/s)=167.6m

- idku

BINGO

- ganeshie8

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

- idku

Yup. Like that \(\Downarrow\) really
(Horizontal displacement) = (\(\color{red}{\rm 2 \times }\) 2.54s) × (33m/s)=167.6m

- idku

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

- idku

Thank you!

- idku

And lets see If I can also bear through question 5.
------------------------------------------
5) What is the speed of the cannonball 3.7 seconds after it was shot?

- idku

I know that
\(\rm t = 3.7(s)-2.54(s)=1.16(s)\)
\(\rm u = 41(m/s)\)
---------------------
I am trying to use
\(\rm v=u+at\)
but I keep arriving on wrong results... lets see...

- idku

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

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