elp!!

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- YumYum247

elp!!

- schrodinger

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- YumYum247

hold up...

- YumYum247

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- YumYum247

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- YumYum247

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- YumYum247

@IrishBoy123 Please check my work!!!! :)

- IrishBoy123

will look at it

- YumYum247

aight thanks :)

- YumYum247

is my method of approaching the problem correct??????/

- 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

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

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

you mean you tried to find the exact angle?

- YumYum247

yes :D

- 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

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

what du i du?

- 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

aight let me try that....:)

- YumYum247

Yes it worked!!!! :"D
Thanks and I Luv your foreverrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr!!!!!!!! :"D

- 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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