Question

A ball is thrown straight down from the top of a 454-foot building with an initial velocity of -18 feet per second. Use the position function below for free-falling objects.
s(t) = -16t2 + v0t + s0
What is its velocity after 3 seconds?
v(3) =
-114 ft/s

What is its velocity after falling 100 feet?
v =

Answer 1

the vol after 3 second= 18 x 3

Answer 2

no, velocity after 3 seconds is not 18 * 3

Answer 3

i already got the V after 3s

Answer 4

Its -114ft/s

Answer 5

yes, agreed.

now, how are you going to find the second part, any ideas?

Answer 6

double differentiate?...lol...i have no idea

Answer 7

if you know the time at which the position of the falling object is 100 feet below the starting point, can't you then calculate velocity?

Answer 8

yeah .. you are right..but how do i get the time?

Answer 9

what will the value of \(s(t)\) be at that point?

Answer 10

8t^2+9t-177

Answer 11

uh, no...

what is \(s(0)=\)

Answer 12

454

Answer 13

we have \[s(t)=-16t^2+v_0(t)+454\]

at the time of interest. the ball has fallen 10p feet from its position at \(s(0)\)

Answer 14

that means \[s(t) = 454-100=354\]right?

Answer 15

right

Answer 16

so solve \[s(t)=354=-16t^2+v_0 t +454\]for \(t\)

Answer 17

then plug the resulting (positive) value of \(t\) into the equation you used to calculate velocity.

Answer 18

I got two ans: 5.27 and -4.14

Answer 19

sounds about right. it is a quadratic, so two solutions, but only the positive one applies to this scenario.

Answer 20

because time is not negative...right?

Answer 21

so 5.27?

Answer 22

yes, we are only using a portion of the parabola in our model of the position.

Answer 23

and what is \{v(5.27)=\]

Answer 24

i had -186.64 as my velocity

Answer 25

sounds plausible. it accelerates at -32 ft/s^2 and has a bit more than 2 more seconds of that acceleration beyond your previous velocity of -114 ft/s.

Answer 26

\[v_f=v_i+at\]

Answer 27

any questions?

Answer 28

lol... it says wrong ans...

Answer 29

i put in 186.64 and then 187

Answer 30

negative sign !

Answer 31

sorry...i did put the negative sign

Answer 32

like -186.64 and -187

Answer 33

hang on a sec, checking your solution for \(t\)

Answer 34

okay

Answer 35

we have \[v_0=-18\text { ft/s}\]is that right?

Answer 36

right

Answer 37

if so, \[354=-16t^2-18t+454\]\[0=-16t^2-18t+100\]\[t=\frac{-(-18)\pm\sqrt{(-18)^2-4(-16)(100)}}{2(-16)}=\]\[\frac{18\pm\sqrt{6724}}{-32}=\frac{18\pm82}{-32}=2,-3.125\]

Answer 38

hold up.. the time is 2s?

Answer 39

yes, \[-16(2)^2-18(2)+45 =-64-36+454=354\]

Answer 40

then the V is gonna be-82?...

Answer 41

sorry, should have checked your value of \(t\) when y0u first got it, but it seemed like it was in the right ballpark.

Answer 42

so then
V= -32(t)-18
V= -32(2)-18
V=-82

Is that right?

Answer 43

yes, just a sec, making a graph for you

Answer 44

yeah...it is!!!
Thanks so much! you are the best!

Answer 45

Here you can see velocity and position changing together, with crosshairs at our two points of interest.

Answer 46

Oh yeah...makes much sense now...

Answer 47

a better one, forgot one of the lines:

Answer 48

Once again thank you very much! you are a lifesaver

Answer 49

Glad I could help. These problems are fun once you get the hang of them! Of course, neglecting air resistance, the answers are maybe not quite realistic all the time, but hey...

Answer 50

yeah..lol I am gonna go off now. Do have a nice day!

Answer 51

And you as well. Time for me to go fix the sweetie's burned out headlight...