Question

The height of an object moving vertically is s(t)=16t^2+80t+245 (feet) at time t seconds. Find the object's maximum height and the time at which it occurs. Discuss domain, find the critical point, and use the first or second derivative test to confirm your answer.

Answer 1

what is s'(t)?

Answer 2

oops, it was supposed to be -16t^2
the derivative would be -32t+80

Answer 3

I suspected :)

Answer 4

right?

Answer 5

at what point in tim is s'(t) = 0?

Answer 6

when t=2.5

Answer 7

that when it reache maximum height! s'(t) is the objects velocity.
now fin s(2.5) , thats the max height

Answer 8

oh!

Answer 9

s"(t) will show the the functions is concave down( if its negative, I bet it is! )

Answer 10

wait, what's the domain?

Answer 11

s''(t)=-32

Answer 12

kinda depends on the model assume that the domain is when the object is higher than 0 so I assume its the values of t such that s(t) > 0

Answer 13

i don't have an actual model, i only have the information that i posted

Answer 14

would 2.5 be the critical point?

Answer 15

yea so it not clear what the domain would be. I the most general sense it all values of t. MATHEMATICALLY. In other words there is a s(t) for all values of t. BUT when s(t) is the object on the ground?? If so the modle isnt accurate when s(t) < 0 .

Answer 16

\[16(t^2+5) + 245\]\[16(t+\frac{5}{2})^2+245-(\frac{5}{2})^2* 16)\]\[16(t+\frac{5}{2})^2+245-100\]\[y= 16(t+\frac{5}{2})^2
+145\]( -5/2,145)

Answer 17

i'm confused?

Answer 18

\[or: x=\frac{-b}{2*a}\]b=80
a=16
substitution x and solve for y

Answer 19

yea, actually Im waiting for an explanation I have the roots of the position functions as 1/4( 10 + sqrt{345}) and 1/4( 10 - sqrt{345})

Answer 20

s(t) > 0 for t between 2.14... and 7.14...

Answer 21

um, how do i show this?

Answer 22

[-2.14, 7.14]

Answer 23

how did you get the 1/4(10+sqrt{345}) and 1/4(10-sqrt{345}?

Answer 24

factoring out the -16 and completing the square. to find the zeros of s(t). BUT I submit that the domain is all values of t because your model does not say that the object can have a negative value.

Answer 25

yes, but it is asking for the maximum. usually maximums are positive, right?

Answer 26

in this case it is s(2.5) is positive. it the maximum height.

Answer 27

345

Answer 28

yes

Answer 29

so yea, your problem does not explicity state that the model ends when s(t) < 0. Another way of looking a t this is assume you are foring this object off a cliff into the Grand Canyon and -5000 is the bottoms then the domain of the MODEL is different. I would not make any assumptions as to the configuration and say that the domain is all values of t.

Hope I dont get you point off with my assumption.

Answer 30

it's ok, but i would like to see most of the steps you used in finding the supposed domain [2.17, 7.14] so I can write it down. please?

Answer 31

its like Nancy did rewrite s(t) like so -16( t^2-5t) = -245 , complete the square and solve for t

Answer 32

ok

Answer 33

good luck! Actually I would repost the question to see if someone else can give some insight to the the meaning of the domain of a position function.

I'll watch with great interest :)

Answer 34

thanks, i also posted it in the group for calculus, but no one has responded to that one yet

Answer 35

There is a Calculus group!?!?! Im there!

Answer 36

there is:)