anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
s(t)=-4.9t+vt+s
anonymous
  • anonymous
hmm are you aloud to use derivatives?
anonymous
  • anonymous
missed a t >.<

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

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