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\[v = {g \over k} (1 - e ^{-kt})\] v = velocity g = gravitational constant which is \[9.8 ms ^{-2}\] k = (measured in \[s^{-1}\] t = seconds what is e? in the equation
lanaaa to the rescue :p
which course are u taking lilg/?

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you know already ee can you help me?
differentiation?
i think it involves differentiation
this is a differential equation and e^-kt is the homogeneous solution
the value for k (s^-1) is 0.15 the value of t you can say is 1 second
what would it be -kte^-kt?
0? :s
I still dont get what ur question is, why did u differentiate that? write down the full question lilg
its really long one minute
Take ur time im not leavin:)
It is already known that objects falling under the influence of gravity obey the equations of constant acceleration, with a gravitational constant g (usually taken to be 9.8 ms ^-2) that applies to all objects independent of their size or mass Under more realistic conditions, objects falling through air will quickly reach a state of constant velocity, this velocity, which depends on the size, shape, and mass of the falling object, is called terminal velocity. you are on a team developing parachutes for skydivers the velocity v (in ms^-1) of a typical skydiver after t seconds is modelled by your team using the equation \[v = {g \over k} (1 - e ^{-kt})\] the value of k (measured in s^-1) is used to model the different stages of the skydivers decent. The table shows some values found through the experimentation: TYPE OF MOTION k(s^-1) Human in free fall 0.15 Parachute deployed 2.00 1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist 2) what is the terminal velocity for the skydiver? theres more questions but yes im stuck on 1 :p
anyone?
do u k anything abt Euler's formula ?
no
e^x=e^z(cosy + i siny) this is the formula
Jop euler cant be used here:S
http://en.wikipedia.org/wiki/Euler%27s_formula
WHY?
e^2.00(1 second) = cos(1) + 2sin(1)
did it work with u?
that would give me -1.223244275 (btw i made a mistake it should -2sin(1))
so \[{9.8 \over 2.00} (-1.223244275)\]
but shoudn't it be \[{9.8^{-2} \over 2.00^{-1}}(-1.223244275)\]
which gives you -0.02547364171
-0.02547364171^-1
is that right?
which is -39.25626384
i think it makes sense But i'm nt sure
but i also forgot to take in account that 2.00 should 2.00^s-1 :S that confuses me the most
Is your question really just what e is?
thats one of my questions but i can see gets taken out using the eulers formula, if thats what im supposed to use in this. but i need help working out question 1 really
if someone can show me how they would write the equation out with the values that would help me understand alot better
http://www.wolframalpha.com/input/?i=y+%3D+10%281+-+e^%28-2t%29%29+from+t%3D0+to+5
where did you get 10 from? and what about when it says k is measured in s^-1?
Okay okay. Well, the question of what e is is not too hard. e is a mathematical constant. Its value is approximately 2.71828
ok thanks for that smoothmath
also g = 9.8 ms^-2
Yup. If you want to read more about it or understand why it's important, wikipedia is a good source, as always. http://en.wikipedia.org/wiki/E_(mathematical_constant)
just putting arbitrary values ... e is Euler's constant ... your velocity should be increasing with constant acceleration ... but since air resistance is directly proportional to velocity .... after certain velocity .. your velocity remains constant ... just another decay equation!!
{lilg scratches his head}
1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist
I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.
what im mainly confused about is when i use these values in the equation FOR EXAMPLE g = 9.8 ms^-2 to put it in the equation as 9.8^-2 or just 9.8
just 9.8 =)
and the same for k when it says it is measured in s^-1 so i should just leave it as 0.15 or 2.00 instead of 0.15^-1
When they say "ms^-2" that's the unit only. \[ms^{-2} = \frac{m}{s^2}\] Which is the saem as "meters per second squared."
also the end result v is in ms^-1 so the end answer e.g. is 20 would i write it as 20^-1 or just 20
Yup. Don't worry too much about the units here =)
ok :)
soooo if you can just check one answer for me if its correct im going to use k = 0.15 and t = 1 second so i would get \[v = {9.8 \over 0.15}(1 + \cos(1) - 0.15\sin(1))\] or is it - cos(1)
Those are units of measurement. If I ask you how much something costs, and you answer "20," I will get angry and ask you "20 WHAT?" Same thing here. Velocity is measured in meters per second, which can be written as \[\frac{m}{s}\]
Oh my god no no no. Ignore that fancy pants crap those other people were giving you. That's just making the problem complicated. Just plug in the right values to the original equation and solve.
can you show me how you would plus in the values in the equation please it will help me understand much better
use the value of k = 0.15 and t = 1 second
are u asking the value of e? it is approximately equal to 2.71..
Sure. k = .15 t=1 \[\frac{g}{k}(1-e^{-kt}) = \frac{9.8}{0.15}(1-e^{-0.15*1})\]
Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)
Still confused?
do i not use eulers formula then?
when i have e^-0.15*1 what do i do with that?
You don't need to, so I don't see why you should.
\[a^{-x} = (\frac{1}{a})^x\]
so would it be \[({1 \over e})^{-0.15}\]
That's the rule for negative exponents. Take the reciprocal and make the exponent positive. Alternatively, you could just put in that expression to a calculator and it'll do the work for you.
Not quite. \[(\frac{1}{e})^{0.15}\]
oh ok yes my mistake
so it would be \[(1 - ({1 \over e})^{0.15})\]
Yessir.
thanks alot mate appreciate your great help
My pleasure =)

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