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lilg132 Group TitleBest ResponseYou've already chosen the best response.0
\[v = {g \over k} (1  e ^{kt})\] v = velocity g = gravitational constant which is \[9.8 ms ^{2}\] k = (measured in \[s^{1}\] t = seconds what is e? in the equation
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
lanaaa to the rescue :p
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
which course are u taking lilg/?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
you know already ee can you help me?
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
differentiation?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
i think it involves differentiation
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
this is a differential equation and e^kt is the homogeneous solution
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
the value for k (s^1) is 0.15 the value of t you can say is 1 second
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
what would it be kte^kt?
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
I still dont get what ur question is, why did u differentiate that? write down the full question lilg
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
its really long one minute
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
Take ur time im not leavin:)
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

JopHP Group TitleBest ResponseYou've already chosen the best response.0
do u k anything abt Euler's formula ?
 2 years ago

JopHP Group TitleBest ResponseYou've already chosen the best response.0
e^x=e^z(cosy + i siny) this is the formula
 2 years ago

lalaly Group TitleBest ResponseYou've already chosen the best response.0
Jop euler cant be used here:S
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Euler%27s_formula
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
e^2.00(1 second) = cos(1) + 2sin(1)
 2 years ago

JopHP Group TitleBest ResponseYou've already chosen the best response.0
did it work with u?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
that would give me 1.223244275 (btw i made a mistake it should 2sin(1))
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
so \[{9.8 \over 2.00} (1.223244275)\]
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
but shoudn't it be \[{9.8^{2} \over 2.00^{1}}(1.223244275)\]
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
which gives you 0.02547364171
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
0.02547364171^1
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
is that right?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
which is 39.25626384
 2 years ago

JopHP Group TitleBest ResponseYou've already chosen the best response.0
i think it makes sense But i'm nt sure
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
but i also forgot to take in account that 2.00 should 2.00^s1 :S that confuses me the most
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Is your question really just what e is?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
if someone can show me how they would write the equation out with the values that would help me understand alot better
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=y+%3D+10%281++e^%282t%29%29+from+t%3D0+to+5
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
where did you get 10 from? and what about when it says k is measured in s^1?
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.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
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks for that smoothmath
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
also g = 9.8 ms^2
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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!!
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
{lilg scratches his head}
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
just 9.8 =)
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.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."
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Yup. Don't worry too much about the units here =)
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.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 \[\frac{m}{s}\]
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
can you show me how you would plus in the values in the equation please it will help me understand much better
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
use the value of k = 0.15 and t = 1 second
 2 years ago

rs32623 Group TitleBest ResponseYou've already chosen the best response.0
are u asking the value of e? it is approximately equal to 2.71..
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Sure. k = .15 t=1 \[\frac{g}{k}(1e^{kt}) = \frac{9.8}{0.15}(1e^{0.15*1})\]
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Still confused?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
do i not use eulers formula then?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
when i have e^0.15*1 what do i do with that?
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
You don't need to, so I don't see why you should.
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
\[a^{x} = (\frac{1}{a})^x\]
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
so would it be \[({1 \over e})^{0.15}\]
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
Not quite. \[(\frac{1}{e})^{0.15}\]
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
oh ok yes my mistake
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
so it would be \[(1  ({1 \over e})^{0.15})\]
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
thanks alot mate appreciate your great help
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.1
My pleasure =)
 2 years ago
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