YumYum247
  • YumYum247
elp!!
Physics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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YumYum247
  • YumYum247
elp!!
Physics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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YumYum247
  • YumYum247
hold up...
YumYum247
  • YumYum247
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YumYum247
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YumYum247
  • YumYum247
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YumYum247
  • YumYum247
@IrishBoy123 Please check my work!!!! :)
IrishBoy123
  • IrishBoy123
will look at it
YumYum247
  • YumYum247
aight thanks :)
YumYum247
  • YumYum247
is my method of approaching the problem correct??????/
IrishBoy123
  • IrishBoy123
absolutely 2 points 1 the question is unclear, are we assuming circle is centred at 42m or starts at 42m? in practise, this doesn't not matter, as the numbers are out by miles, but you should spot this when you are drawing the thing. perhaps you did. 2 yes, i agree with conclusions, but a/ your rounding is out for starters. 34 cos(18) = 32.33 b/ you used the quadratic formula when you had something simpler in your answer, you're using vector arrows and stuff that doesn't look right to me, so if we just call vertical displacement y(t) then you have \(y(t) = 10.5 t + \frac{1}{2}(-9.8)t^2 = t (10.5 - \frac{1}{2}(9.8)t)\) so \(y = 0 \implies t = 0 , \) or \(10.5 - \frac{1}{2}(9.8)t = 0\) so that solves to t = 2.14, fine, but I get 69.27, ie 69.3 for the distance it lands. small potatoes, but this is from the rounding. IOW fine but you could tighten it up a bit not sure what the third page is about.
IrishBoy123
  • IrishBoy123
as for the waffly bit, i'd personally think a/ change elevation of cannon b/ change velocity of launch c/ change distance between cannon and targe you point about changing elevation to above 45 shows a good understanding and is well made other solutions, eg raising/lowering the cannon, making the target or the balloon massive, asking your deity to change the value of gravity, they all work too.....but not sure about that and i don't know your instructor yep, well done
YumYum247
  • YumYum247
but i actually went with manipulating the angle from 18deg( given) to a different range of angles from 78 to 70 to even 80 but the thing is that none of them quite land the balloon right on the target....i've about 1 to 2 meters short.... :"(
IrishBoy123
  • IrishBoy123
you mean you tried to find the exact angle?
YumYum247
  • YumYum247
yes :D
YumYum247
  • YumYum247
should i try to manipulate the vertical displacement of the launcher and keep the angle of the cannon the same 18deg...would that work?
YumYum247
  • YumYum247
but that would cause the balloon is gain more distance on the horizontal axis and my balloon will end up going way over my mark....42m :(
YumYum247
  • YumYum247
what du i du?
IrishBoy123
  • IrishBoy123
do what you just did but be a bit more general if we use x and y for 'across' and 'up' then \(x(t) = 34 cos \theta \ \times \ t\) where \(\theta\) is our magic angle if it reaches \(x = 42\) at time \(t = T\), we can say \(T =\frac{42}{34 cos \ \theta} \) also we can say that \(y(T) = 0 \) because it has landed so \(0 = (34 sin \ \theta \ T - \frac{1}{2}gT^2) = 34 sin \ \theta - \frac{1}{2}gT = 0\) so \(34 sin \ \theta = \frac{1}{2}gT = \frac{1}{2}(9.8)\frac{42}{34 \ cos \ \theta}\) so \(2 sin \ \theta \ cos \ \theta = 2 \frac{1}{2}(9.8)\frac{42}{34^2 }\) the LHS is also \(sin \ 2 \theta\), this is a double angle identity that gives \(sin \ 2\theta = 0.356\) \(\theta = 10.43^o\) try that
YumYum247
  • YumYum247
aight let me try that....:)
YumYum247
  • YumYum247
Yes it worked!!!! :"D Thanks and I Luv your foreverrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr!!!!!!!! :"D
IrishBoy123
  • IrishBoy123
the *2* solutions, as you correctly identified there would be, are \(sin 2θ=0.356 \implies 2 \theta = 20.86^o, \ 180^o - 20.86^o \implies \theta = 10.43^o, \ 79.57^o \) there are rounding errors in there, now

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