Question

I would like to run my answer by somebody. I am not too confident in it.

Answer 1

This is the problem.

Answer 2

I have figured out the answer to be 100/83

Answer 3

\[s(2)=16(2)^2+100−16(2)^2+100÷2−(−16)(2)^2+100 \]

Answer 4

I simplified the equation as much as possible and received 100/83

Answer 5

First, the 100 should subtract off
Also, I think you should put in 2-h for t in
\[ \frac{s(2)-s(2-h)}{2-(2-h)} \]
s(2)= -16*4+100=-36
s(2-h)= -16*(2-h)^2 + 100 = -16(4-2h+h^2) +100= 36+32h-36h^2
Now subtract s(2)-s(2-h)
you should get -32h+36h^2
the denominator 2-2+h= h so your fraction becomes
\[ \frac{-32h + 36h^2}{h} \]
or -32 + 36h
now let h-->0 (and 2-h --> 2)

Answer 6

You still have to figure out the units, but they should be ft/sec

Answer 7

*typo on s(2)= +36 ft

Answer 8

I'm totally lost on this explanation

Answer 9

The velocity at t = 2 s should be -64 ft/s. That limit is another variation of the derivative. \[s(t) = -16t^2 + 100\]\[v(t) = -32t \rightarrow v(2) = -64 \space ft/s\]

Answer 10

Thank you. I figured it out though. I feel more confident knowing that someone else got the same answer.

Answer 11

Oh, okay, good =)