Math-Based-Physics question.

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Math-Based-Physics question.

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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. -------------------------------------------------------------
STOP IT, GOD DAMN IT !
xD
`PART 1:` 1) What is the initial speed of the cannonball? \(\sqrt{33^2(m/s)+25^2(m/s)~}=41(m/s)\)
`PART 2:` 2) What is the initial angle θ of the cannonball with respect to the ground?`PART 1:` \(\theta=\tan^{-1}\left(25/33\right)\approx37.15\)
`PART 3` 3) What is the maximum height the cannonball goes above the ground?
\(\Large v^2=u^2+2ax\) (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...
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
\(\Large v^2=u^2+2ax\) (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!
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?
\(\Large x = ut + \dfrac{1}{2}at^2\) \(\Large 31.8 = 25t - \dfrac{9.81}{2}t^2\) t = 2.44229 t = 2.65455 then I can use: Horizontal displacement = time × initial horizontal velocity my problem now is which time to take
you should not get multiple "t" values the vertex is a "single" point |dw:1441337672713:dw|
31.855249745=25t−(9.81/2)t^2
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
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
(Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = (2.54s) × (33m/s)=83.82s but the practice thingy tells me that is wrong
oh wait 2.54 is half of the whole time, right?
Yup. (Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = \(\color{red}{\rm 2 \times }\) (2.54s) × (33m/s)=167.6m
BINGO
Yes, 2.54 = time at which the particle reaches max height
Yup. Like that \(\Downarrow\) really (Horizontal displacement) = (\(\color{red}{\rm 2 \times }\) 2.54s) × (33m/s)=167.6m
Yes, so from max back to v=0 another 2.54s
Thank you!
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?
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...
oh, I also tried t=3.7 without subtracting and it was still wrong

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