Question

the model for an object falling is given by v(t) = (32.2/k) - (32.2/k)(e^-kt) solve k so it takes 2.6 seconds for the object to hit the ground

Answer 1

The formula is
\[v(t)=\frac{32.3}{k}-\frac{32.3}{k}(e^{-kt})\]
Correct?

Answer 2

exactly

Answer 3

Hehe cool. So then our \(t\)-value is 2.6 and assuming \(v(t)\) Is the height of the object from the ground, then we have \(v(t)=0\). Is \(v(t)\) the height of the object from the ground?

Answer 4

It says that it falls from 100 feet so I'm assuming so.

Answer 5

Okay. So we can make the formula:
\[0=\frac{32.3}{k}-\frac{32.3}{k}(e^{-2.6k})\]
And we must isolate for \(k\).

Answer 6

Wouldn't v(t) be 2.6 since we are substituting 2.6 for t?

Answer 7

No...because \(t=2.6\) does not mean \(v(t)=2.6\)

Answer 8

Ok understood

Answer 9

I get 13.46 for K

Answer 10

So...we can simplify a little bit.
\[0=\frac{32.3}{k}(1-e^{-2.6k})\]
Since we know that \(\frac{32.3}{k}\) Will never be zero, then, through the property of zero, we know that
\[1-e^{-2.6k}=0\]
And we can isolate for k here
\[e^{-2.6k}=1\]
\[-2.6k=ln(1)\]...Ermmm wait no something is wrong. I believe that I substituted something in the wrong place...Haha my bad. Something is wrong with this...ill end up with k=0 and that's wrong.
I feel like there is something more to this question. I wish I could figure it out, but I must go. Ill look at this tomorrow at a similar time.