Question

A paper clip is dropped from the top of a 144-ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = -16t^2 + 144. What is its velocity in ft/sec when it hits the ground? (2 points)

-96
-64
-32
0

I think it's C

Answer 1

Why C?

Can you share your works/steps, please?

Answer 2

http://openstudy.com/updates/584d6affe4b071fda882243d

Answer 3

cuz when you derivative the equation you get -32

Answer 4

What about this "144-ft tower" and this "initial velocity of 16 ft/sec"?

Answer 5

... no idea ._.

Answer 6

Let's proceed!!!

Answer 7

kay x3

Answer 8

Of course you have studied these three laws of motion!

Answer 9

yes

Answer 10

That is awesome!

so as I am used to do ,,, We need to use what we have to get what we need

- A paper clip is dropped from the top of a 144-ft tower,
so it is dropped from the top (falls with the gravitational acceleration \(\Large g(9.81m/s^2)\) + from the top of a 144-ft tower (falls from a vertical distance equal to the height of the tower)

- "with an initial velocity of 16 ft/sec" i.e it started to move not from stand still, but it has an initial speed of \(\Large v_i=16 ft/sec\)

- "....when it hits the ground" means two things
* it has traveled all the tower height
* it has no velocity at that point, i.e its final speed is zero \(\Large v_f=0\)

Answer 11

According to these data, in addition to the required "What is its velocity in ft/sec when it hits the ground", we are going to use which equation of these I sent you?

Answer 12

I must admit I'm puzzled by this question. @emar Why do you say that the final velocity is 0. I'm not saying you are wrong.

Answer 13

I have to go right now. I#ll be back later.

Answer 14

uh we use the 3rd one?

Answer 15

Which is?

Answer 16

\[V_1^2=v_o^2+2as\]

Answer 17

You hit the right target!

Can you plug in the given data into?

Answer 18

uuhhhh i'm not sure about this but
\[V_1^2=v_0^2+2(144)(16)\]

Answer 19

and what about \(v_o\) ?

Answer 20

uummm, idk :\

Answer 21

close - v_0 is your initial velocity, and your acceleration isn't 16... it's the SECOND derivative of the position function s(t) = -16t^2 + 144

Answer 22

Nice note, @inkyvoyd

so what would you say about that approach!?

Answer 23

final velocity is not 0

it asks for velocity 'when it hits ground' - but this means in the instant before it hits the ground.

Answer 24

This is what I have corrected in the picture!

Answer 25

When it hits the ground its position is 144 ft

so use the first equation to work out t

then do the derivitive to get the velocit function and put the value obatained for t into that derivitive to get v

Answer 26

Soz - actually - from the equation - its position is 0 when it hits the ground

Answer 27

"When it hits the ground its position is 144 ft"

I don't think so....because its position is zero at hitting the ground...look at the axes and coordinates!

Answer 28

"Soz"???????

Answer 29

From "The Urban Dictionary" :-)
"soz." Nonsensical internet slang term for "sorry", used by illiterate morons who for some reason substitute a “z” for “rry”,

Answer 30

and for the question itself....did you get my pic?

Answer 31

yes - but the equation given is
s(t) = -16t^2 + 144

so v(t) = -32t

v(3) = -96 fps

You have used the 'standard' equation rather than that given

(I agree that yours is the normal answer - but it is not as given - I think th equestion is incorrect given that the initial vleocity is not 0)

Answer 32

the question IS incorrect as
v(0) = 0
but it tell us th at V0 =-16

Answer 33

you are very precise and have a nice overall view of the case!
I thank you in person as you have illustrated some points that were not clear for me

Thanks a lot..

I though so that the problem is not completely stable, as how does the initial velocity have a value in the mean time the final velocity should not exceed 96!!!!

That made me confused!

so thanks again...

Answer 34

@volpina should be told that this problem is not quite having a correct data, although we did our best to analyze it logically..

Answer 35

I think (s)he left the forum...

Answer 36

Never mind...but I am glad that I have found someone is confused like me ;))

Answer 37

Yeah the question is confusing at first sight. That's why I was puzzled by it.

Answer 38

Yes, @welshfella you were right!

But when I involved, I found that it needs to treat the problem when dropping according to the position equation S(t), and it actually conflict with the given data (like initial speed)

Answer 39

exactly

Answer 40

I knew there was something wrong but I couldn't quite see what it was.

Answer 41

thanks for @MrNood