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December 20, 2014

December 20, 2014

Posted by **Sarah** on Monday, June 9, 2014 at 12:16am.

- trig -
**Reiny**, Monday, June 9, 2014 at 9:00amlet's fit the data to a curve of the type

f(t) = a cos k(t +d) + c

from highest to lowest = 65-25= 40 cm

so a = 20

from highest to lowest = .3 s to 1.8 s

so period = 3 seconds

2π/k = 3

k = 2π/3

so far we have

f(t) = 20 cos (2π/3) (t +d) + c

let's raise the curve 45 cm

f(t) = 20 cos (2π/3) (t +d) + 45

and finally , our max now occurs at t = 0, but we want to move our curve .3 to the right, so the max occurs at t=.3

f(t) = 20 cos (2π/3) (t - .3) + 45

so when does f(t) = 50 ?

20 cos (2π/3)(t - .3) + 45 = 50

cos (2π/3)(t - .3) = .25

so (2π/3)(t-.3) = 1.318116

t - .3 = .62935..

**t = .92935...**

or (2π/3)(t-.3) = 2π - 1.318116

**t= 2.6706....**

of course since the period is 3 seconds, adding or subtracting 3 seconds from any solution will yield a new solution

so how about .92935 + 3 = 3.92935 seconds ??

So the next two times after 2 seconds to reach a height of 50 cm is

t = 2.6706 and t = 3.92935

check:

f(2.6706...) = 50

f(3.92935...) = 50 , used my calculator.

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