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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?
 2 years ago
 2 years ago
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?
 2 years ago
 2 years ago

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AwstinfBest ResponseYou've already chosen the best response.1
hmm are you aloud to use derivatives?
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
4.9t^2 I think you mean :)
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
and yes i can use derivatives
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
im just not sure how to use the information given
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
im so confused right now
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
this rate of change thing is going to do me in
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
hmm so vo and so are both unknown?
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
yes but i am assuming s0 would = 0 at 5.6 seconds
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
yeah thats right the vo is just the pain in the a part of the problem
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
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
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
that seems like it might be too easy though
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
we dont drop things with an intial velocity though lol, thats against the definition of dropping lol
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
I am so lost on any problems dealing with rate of change
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
I may have to drop this course if I do not pick these up soon :(
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
4.9(5.6^2) should be your answer assuming that dropping means letting go from rest
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
I am having a hard time telling where the information they give me plugs in
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
Say they give me a distance of 108 meters, I assume thats s
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
if they give me a velocity that v0t
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
and a measurement of time = t
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
so if they give me t and vot
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
and solve for the value of t
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
but when they give me t and s im lost lol
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
and when they only give me t i am more than lost
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
no that more like my next problem, but if i were given that question
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
Here ill read type you a problem from my book and try to help clarify things
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
i do not know what to do
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
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?
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
i have no idea how to solve that
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
But they are questions like that correct?
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
I'll show you how to appoarch it if it is
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
(this is straight from my change of rate section)
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
hmmmm on my original question
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
s(t)=4.9+vt+s v(t)=s'(t)
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
so 9.8t+v would be the derivative
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
would that be the velocity?
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
thats your impact velocity
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
ok so when s=0 Velocity = 54.88?
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
now i confused myself lol
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
hahaha remember the derivative of the distance equation is the velocity at time equation
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
at 5.6 secs s=0 since thats impact right?
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
wait nno no s0 is not o
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
so at time t=5.6s s would be 0 not s0
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
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.
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
well i am trying to understand
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
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
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
4.9t^2+vt+s 9.8t+v is the derivative correct?
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
with that i can get the velocity at impact
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
yes but you need to know what v initial is
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
well wouldnt that be as simple as making t 0?
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
I dont think you can do that because you are stating that v(5.6)=0
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
Try reposting this in the physics section, they are really good at explain this kind of stuff
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
http://www.blurtit.com/q3330030.html
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
the initial velocity is 0
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
Sorry I was trying to be ambiguous I didnt want to give you improper information
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
I am in your same level class lol
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
I just do about 6 hours of hw a day
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
yeah it blows I have no social life atm just study study
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
com 101 is more time consuming than i thought
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
:( 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
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
chem, com 101, calc and us history
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
history and chem is ok, but com 101 is taking way to much of my time
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
taking way to much time writing and preparing for speeches
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
Thats why I stick to courses that strictly deal with numbers
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
lol its a req for my degree or i would never touch it
 2 years ago

AwstinfBest ResponseYou've already chosen the best response.1
:( I just wanna take comparitive virology already, all of these courses building up to it are just BS
 2 years ago
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