ChrisV
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?
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ChrisV
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s(t)=-4.9t+vt+s
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hmm are you aloud to use derivatives?
ChrisV
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missed a t >.<
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-4.9t^2 I think you mean :)
ChrisV
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yea
ChrisV
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-4.9t^2 +vt+s
ChrisV
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and yes i can use derivatives
ChrisV
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im just not sure how to use the information given
ChrisV
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im so confused right now
ChrisV
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this rate of change thing is going to do me in
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hmm so vo and so are both unknown?
ChrisV
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yes but i am assuming s0 would = 0 at 5.6 seconds
ChrisV
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since thats impact
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yeah thats right the vo is just the pain in the a part of the problem
Awstinf
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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
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that seems like it might be too easy though
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we dont drop things with an intial velocity though lol, thats against the definition of dropping lol
ChrisV
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I am so lost on any problems dealing with rate of change
ChrisV
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I may have to drop this course if I do not pick these up soon :(
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:(
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4.9(5.6^2) should be your answer assuming that dropping means letting go from rest
ChrisV
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I am having a hard time telling where the information they give me plugs in
ChrisV
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Say they give me a distance of 108 meters, I assume thats s
ChrisV
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if they give me a velocity that v0t
ChrisV
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and a measurement of time = t
ChrisV
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so if they give me t and vot
ChrisV
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i find the derivative
ChrisV
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and solve for the value of t
ChrisV
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but when they give me t and s im lost lol
Awstinf
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s(t)= -4.9t^2+v0t+108?
ChrisV
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and when they only give me t i am more than lost
ChrisV
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no that more like my next problem, but if i were given that question
Awstinf
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Here ill read type you a problem from my book and try to help clarify things
ChrisV
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i do not know what to do
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read-type*
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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?
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questions like that?
ChrisV
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i have no idea how to solve that
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But they are questions like that correct?
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I'll show you how to appoarch it if it is
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(this is straight from my change of rate section)
ChrisV
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hmmmm on my original question
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yeah?
ChrisV
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s(t)=-4.9+vt+s
v(t)=s'(t)
ChrisV
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oops 4.9t^2
ChrisV
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so -9.8t+v would be the derivative
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yeah
ChrisV
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-9.8(5.6)+v=0
v= 54.88
ChrisV
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would that be the velocity?
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thats your impact velocity
ChrisV
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ok so when s=0 Velocity = 54.88?
Awstinf
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assuming that v0=0 yes
ChrisV
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now i confused myself lol
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hahaha remember the derivative of the distance equation is the velocity at time equation
ChrisV
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-4.9t^2 +vt+s
ChrisV
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well if s=distance
ChrisV
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at 5.6 secs s=0 since thats impact right?
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yes
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wait nno no s0 is not o
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0*
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here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s
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so at time t=5.6s s would be 0 not s0
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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.
ChrisV
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hehe
ChrisV
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well i am trying to understand
Awstinf
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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
ChrisV
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-4.9t^2+vt+s
9.8t+v is the derivative correct?
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yes sir
ChrisV
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with that i can get the velocity at impact
ChrisV
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-9.8(5.6)+v=0
v= 54.88
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yes but you need to know what v initial is
ChrisV
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well wouldnt that be as simple as making t 0?
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I dont think you can do that because you are stating that v(5.6)=0
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Its hard to explain
ChrisV
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yea i see lol
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Try reposting this in the physics section, they are really good at explain this kind of stuff
ChrisV
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will do thanks
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I was right :)
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the initial velocity is 0
ChrisV
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:)
ChrisV
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:)
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Sorry I was trying to be ambiguous I didnt want to give you improper information
Awstinf
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I am in your same level class lol
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I just do about 6 hours of hw a day
ChrisV
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i took too many courses
ChrisV
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falling behind
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yeah it blows I have no social life atm just study study
ChrisV
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com 101 is more time consuming than i thought
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:( 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
ChrisV
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chem, com 101, calc and us history
ChrisV
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history and chem is ok, but com 101 is taking way to much of my time
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us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do
ChrisV
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taking way to much time writing and preparing for speeches
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Thats why I stick to courses that strictly deal with numbers
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lol
ChrisV
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lol its a req for my degree or i would never touch it
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:( I just wanna take comparitive virology already, all of these courses building up to it are just BS