idku
  • idku
Math-Based-Physics question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
yolomcswagginsggg
  • yolomcswagginsggg
9 + 10?
anonymous
  • anonymous
19
yolomcswagginsggg
  • yolomcswagginsggg
21.

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idku
  • 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. -------------------------------------------------------------
idku
  • idku
STOP IT, GOD DAMN IT !
yolomcswagginsggg
  • yolomcswagginsggg
xD
idku
  • 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
  • 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
  • idku
`PART 3` 3) What is the maximum height the cannonball goes above the ground?
idku
  • 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
  • idku
I just want to do it a plain way using some familiar formula
idku
  • idku
oh, it should be 25 not 33
idku
  • idku
because we are computing vertical displacement we must use vertical velocity
idku
  • 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
  • idku
Okay, I am getting through it with a painful struggle-:(:)
ganeshie8
  • ganeshie8
good, nothing to feel bad about these things take time
idku
  • idku
Like nothing else in this world.... :) But, true that
idku
  • idku
Well, the fact that my teacher is awful gives me more credit
idku
  • idku
4) How far from where it was shot will the cannonball land?
idku
  • 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
  • ganeshie8
you should not get multiple "t" values the vertex is a "single" point |dw:1441337672713:dw|
idku
  • idku
31.855249745=25t−(9.81/2)t^2
idku
  • 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
  • ganeshie8
use the exact value for \(x\), then u will get a single \(t\ value
ganeshie8
  • ganeshie8
use the exact value for \(x\), then u will get a single \(t\) value
idku
  • idku
well, I got a single approximation for t (as for now because I increased my value-precision)
idku
  • idku
2.54 seconds
ganeshie8
  • ganeshie8
looks fine
idku
  • 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
  • idku
oh wait 2.54 is half of the whole time, right?
idku
  • idku
Yup. (Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = \(\color{red}{\rm 2 \times }\) (2.54s) × (33m/s)=167.6m
idku
  • idku
BINGO
ganeshie8
  • ganeshie8
Yes, 2.54 = time at which the particle reaches max height
idku
  • idku
Yup. Like that \(\Downarrow\) really (Horizontal displacement) = (\(\color{red}{\rm 2 \times }\) 2.54s) × (33m/s)=167.6m
idku
  • idku
Yes, so from max back to v=0 another 2.54s
idku
  • idku
Thank you!
idku
  • 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
  • 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
  • idku
oh, I also tried t=3.7 without subtracting and it was still wrong

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