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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?

Mathematics
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s(t)=-4.9t+vt+s
hmm are you aloud to use derivatives?
missed a t >.<

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Other answers:

-4.9t^2 I think you mean :)
yea
-4.9t^2 +vt+s
and yes i can use derivatives
im just not sure how to use the information given
im so confused right now
this rate of change thing is going to do me in
hmm so vo and so are both unknown?
yes but i am assuming s0 would = 0 at 5.6 seconds
since thats impact
yeah thats right the vo is just the pain in the a part of the problem
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
that seems like it might be too easy though
we dont drop things with an intial velocity though lol, thats against the definition of dropping lol
I am so lost on any problems dealing with rate of change
I may have to drop this course if I do not pick these up soon :(
:(
4.9(5.6^2) should be your answer assuming that dropping means letting go from rest
I am having a hard time telling where the information they give me plugs in
Say they give me a distance of 108 meters, I assume thats s
if they give me a velocity that v0t
and a measurement of time = t
so if they give me t and vot
i find the derivative
and solve for the value of t
but when they give me t and s im lost lol
s(t)= -4.9t^2+v0t+108?
and when they only give me t i am more than lost
no that more like my next problem, but if i were given that question
Here ill read type you a problem from my book and try to help clarify things
i do not know what to do
read-type*
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?
questions like that?
i have no idea how to solve that
But they are questions like that correct?
I'll show you how to appoarch it if it is
(this is straight from my change of rate section)
hmmmm on my original question
yeah?
s(t)=-4.9+vt+s v(t)=s'(t)
oops 4.9t^2
so -9.8t+v would be the derivative
yeah
-9.8(5.6)+v=0 v= 54.88
would that be the velocity?
thats your impact velocity
ok so when s=0 Velocity = 54.88?
assuming that v0=0 yes
now i confused myself lol
hahaha remember the derivative of the distance equation is the velocity at time equation
-4.9t^2 +vt+s
well if s=distance
at 5.6 secs s=0 since thats impact right?
yes
wait nno no s0 is not o
0*
here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s
so at time t=5.6s s would be 0 not s0
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.
hehe
well i am trying to understand
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
-4.9t^2+vt+s 9.8t+v is the derivative correct?
yes sir
with that i can get the velocity at impact
-9.8(5.6)+v=0 v= 54.88
yes but you need to know what v initial is
well wouldnt that be as simple as making t 0?
I dont think you can do that because you are stating that v(5.6)=0
Its hard to explain
yea i see lol
Try reposting this in the physics section, they are really good at explain this kind of stuff
http://www.blurtit.com/q3330030.html
will do thanks
I was right :)
the initial velocity is 0
:)
:)
Sorry I was trying to be ambiguous I didnt want to give you improper information
I am in your same level class lol
I just do about 6 hours of hw a day
i took too many courses
falling behind
yeah it blows I have no social life atm just study study
com 101 is more time consuming than i thought
:( 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
chem, com 101, calc and us history
history and chem is ok, but com 101 is taking way to much of my time
us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do
taking way to much time writing and preparing for speeches
Thats why I stick to courses that strictly deal with numbers
lol
lol its a req for my degree or i would never touch it
:( I just wanna take comparitive virology already, all of these courses building up to it are just BS

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