Question

s(t)=-4.9+vt+s

To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 5.6 seconds after the stone is dropped?

Answer 1

s(t)=-4.9t+vt+s

Answer 2

hmm are you aloud to use derivatives?

Answer 3

missed a t >.<

Answer 4

-4.9t^2 I think you mean :)

Answer 5

yea

Answer 6

-4.9t^2 +vt+s

Answer 7

and yes i can use derivatives

Answer 8

im just not sure how to use the information given

Answer 9

im so confused right now

Answer 10

this rate of change thing is going to do me in

Answer 11

hmm so vo and so are both unknown?

Answer 12

yes but i am assuming s0 would = 0 at 5.6 seconds

Answer 13

since thats impact

Answer 14

yeah thats right the vo is just the pain in the a part of the problem

Answer 15

wait wait wait, it seems to be lacking the classic physics wording of it was dropped with an unknown intial velocity, it being dropped and not thrown I would maybe go as far to say the the intial velocity is 0

Answer 16

that seems like it might be too easy though

Answer 17

we dont drop things with an intial velocity though lol, thats against the definition of dropping lol

Answer 18

I am so lost on any problems dealing with rate of change

Answer 19

I may have to drop this course if I do not pick these up soon :(

Answer 20

:(

Answer 21

4.9(5.6^2) should be your answer assuming that dropping means letting go from rest

Answer 22

I am having a hard time telling where the information they give me plugs in

Answer 23

Say they give me a distance of 108 meters, I assume thats s

Answer 24

if they give me a velocity that v0t

Answer 25

and a measurement of time = t

Answer 26

so if they give me t and vot

Answer 27

i find the derivative

Answer 28

and solve for the value of t

Answer 29

but when they give me t and s im lost lol

Answer 30

s(t)= -4.9t^2+v0t+108?

Answer 31

and when they only give me t i am more than lost

Answer 32

no that more like my next problem, but if i were given that question

Answer 33

Here ill read type you a problem from my book and try to help clarify things

Answer 34

i do not know what to do

Answer 35

read-type*

Answer 36

Hokay so, A person standing at the edge of a cliff throws a rock directly upward. it is observed that 2 seconds later the rock is at its maximum height and that 5 seconds after that, it hits the ground at the base of the cliff?

Answer 37

questions like that?

Answer 38

i have no idea how to solve that

Answer 39

But they are questions like that correct?

Answer 40

I'll show you how to appoarch it if it is

Answer 41

(this is straight from my change of rate section)

Answer 42

hmmmm on my original question

Answer 43

yeah?

Answer 44

s(t)=-4.9+vt+s
v(t)=s'(t)

Answer 45

oops 4.9t^2

Answer 46

so -9.8t+v would be the derivative

Answer 47

yeah

Answer 48

-9.8(5.6)+v=0
v= 54.88

Answer 49

would that be the velocity?

Answer 50

thats your impact velocity

Answer 51

ok so when s=0 Velocity = 54.88?

Answer 52

assuming that v0=0 yes

Answer 53

now i confused myself lol

Answer 54

hahaha remember the derivative of the distance equation is the velocity at time equation

Answer 55

-4.9t^2 +vt+s

Answer 56

well if s=distance

Answer 57

at 5.6 secs s=0 since thats impact right?

Answer 58

yes

Answer 59

wait nno no s0 is not o

Answer 60

0*

Answer 61

here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s

Answer 62

so at time t=5.6s s would be 0 not s0

Answer 63

other wise you would get a negative height, it really doesnt matter which way you do it but you should do it right when you are first doing it.

Answer 64

hehe

Answer 65

well i am trying to understand

Answer 66

I'm sorry I really havent been too much help but that question is horribly worded. If I were in your shoes I would assume that the initial velocity is 0 as progress from there. As long as you understand the relationships between the equations, it will make solving these very simple. Drawing whats going on helps

Answer 67

-4.9t^2+vt+s
9.8t+v is the derivative correct?

Answer 68

yes sir

Answer 69

with that i can get the velocity at impact

Answer 70

-9.8(5.6)+v=0
v= 54.88

Answer 71

yes but you need to know what v initial is

Answer 72

well wouldnt that be as simple as making t 0?

Answer 73

I dont think you can do that because you are stating that v(5.6)=0

Answer 74

Its hard to explain

Answer 75

yea i see lol

Answer 76

Try reposting this in the physics section, they are really good at explain this kind of stuff

Answer 77

http://www.blurtit.com/q3330030.html

Answer 78

will do thanks

Answer 79

I was right :)

Answer 80

the initial velocity is 0

Answer 81

:)

Answer 82

:)

Answer 83

Sorry I was trying to be ambiguous I didnt want to give you improper information

Answer 84

I am in your same level class lol

Answer 85

I just do about 6 hours of hw a day

Answer 86

i took too many courses

Answer 87

falling behind

Answer 88

yeah it blows I have no social life atm just study study

Answer 89

com 101 is more time consuming than i thought

Answer 90

:( yeah thats the draw back of underestimating courses. I'm taking physics, chem, physics lab, calc and I really thought this semester would be a tad less stressful

Answer 91

chem, com 101, calc and us history

Answer 92

history and chem is ok, but com 101 is taking way to much of my time

Answer 93

us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do

Answer 94

taking way to much time writing and preparing for speeches

Answer 95

Thats why I stick to courses that strictly deal with numbers

Answer 96

lol

Answer 97

lol its a req for my degree or i would never touch it

Answer 98

:( I just wanna take comparitive virology already, all of these courses building up to it are just BS