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idku
 one year ago
MathBasedPhysics question.
idku
 one year ago
MathBasedPhysics question.

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idku
 one year ago
Best ResponseYou've already chosen the best response.1A 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
 one year ago
Best ResponseYou've already chosen the best response.1`PART 1:` 1) What is the initial speed of the cannonball? \(\sqrt{33^2(m/s)+25^2(m/s)~}=41(m/s)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1`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
 one year ago
Best ResponseYou've already chosen the best response.1`PART 3` 3) What is the maximum height the cannonball goes above the ground?

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large v^2=u^2+2ax\) (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
 one year ago
Best ResponseYou've already chosen the best response.1I just want to do it a plain way using some familiar formula

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, it should be 25 not 33

idku
 one year ago
Best ResponseYou've already chosen the best response.1because we are computing vertical displacement we must use vertical velocity

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large v^2=u^2+2ax\) (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
 one year ago
Best ResponseYou've already chosen the best response.1Okay, I am getting through it with a painful struggle:(:)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1good, nothing to feel bad about these things take time

idku
 one year ago
Best ResponseYou've already chosen the best response.1Like nothing else in this world.... :) But, true that

idku
 one year ago
Best ResponseYou've already chosen the best response.1Well, the fact that my teacher is awful gives me more credit

idku
 one year ago
Best ResponseYou've already chosen the best response.14) How far from where it was shot will the cannonball land?

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large x = ut + \dfrac{1}{2}at^2\) \(\Large 31.8 = 25t  \dfrac{9.81}{2}t^2\) 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
 one year ago
Best ResponseYou've already chosen the best response.1you should not get multiple "t" values the vertex is a "single" point dw:1441337672713:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.131.855249745=25t−(9.81/2)t^2

idku
 one year ago
Best ResponseYou've already chosen the best response.1then 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
 one year ago
Best ResponseYou've already chosen the best response.1use the exact value for \(x\), then u will get a single \(t\ value

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1use the exact value for \(x\), then u will get a single \(t\) value

idku
 one year ago
Best ResponseYou've already chosen the best response.1well, I got a single approximation for t (as for now because I increased my valueprecision)

idku
 one year ago
Best ResponseYou've already chosen the best response.1(Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = (2.54s) × (33m/s)=83.82s but the practice thingy tells me that is wrong

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh wait 2.54 is half of the whole time, right?

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yup. (Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = \(\color{red}{\rm 2 \times }\) (2.54s) × (33m/s)=167.6m

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 2.54 = time at which the particle reaches max height

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yup. Like that \(\Downarrow\) really (Horizontal displacement) = (\(\color{red}{\rm 2 \times }\) 2.54s) × (33m/s)=167.6m

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yes, so from max back to v=0 another 2.54s

idku
 one year ago
Best ResponseYou've already chosen the best response.1And 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1oh, I also tried t=3.7 without subtracting and it was still wrong
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