anonymous
  • anonymous
HELP PLEASE!!!!!!!!!! If a rock is thrown upward with an initial velocity of 64 feet per second from the top of a 25-foot building, write the height (s) equation using this information.
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
use h=ut -1/2 *g*t^2 here u =64ft/sec and g =32 ft/sec^2 for finding when it reaches ground for h plug -25 ft
anonymous
  • anonymous
so its not \[s=-16t ^{2}+v _{0}t+s _{0}\]
anonymous
  • anonymous
its same thing with

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
with S0 as 25ft
anonymous
  • anonymous
what does the \[v _{0} s _{0} \] mean
anonymous
  • anonymous
its v0*t here V0 is initaial velocity and here it is 64 ft/s
anonymous
  • anonymous
so the equation would be S=-16t^2+64(1)+25
anonymous
  • anonymous
yup but with t S=-16t^2+64(t)+25
anonymous
  • anonymous
ok so if I want to find out how high the rock is after 10 sec I would plug it in like this S=-16(10)^2+64(10)+25
anonymous
  • anonymous
yes
anonymous
  • anonymous
also to clear it up the subscript 0 is usually prenounced "knot" and it's basically the starting point of any physics problem, and then you number them 1,2,3,etc there on. You could essentially call it v1 but your next point would be v2. s in physics is used for displacement aka distance, so in your case s0 means the initial height before being thrown
anonymous
  • anonymous
If you have to take physics eventually you split everything into two components the y direction and x direction so you get something like this \[s_{y}=\frac{1}{2}at^2+v_{0y}t+s_{0y}\] which would be your problem since throwing an object vertically only has distance in the y direction
anonymous
  • anonymous
Post such questions in physics group.. also refer to: http://openstudy.com/code-of-conduct
anonymous
  • anonymous
This is a math related subject. as you can solve it through derivatives and calculus. And I find it very annoying of you trying to pretend like a moderator. Let them do their job. Posting a question that is off topic for a particular group (i.e. posting a physics question in the Mathematics group)---- This is solely any type of physics that isn't related to mathematics such as asking quantum physic/fluid dynamics in a group other than physics. This however is math because it's using a quadratic equation involving t (time). Not only that but Calculus is used to solve this also. For example so that you can quit citing the CoC everytime you see displacement problems \[a=-32ft/sec\] \[\int a dt=\int =-32dt=-32t+c=v(t) \] \[v(0)=64 ft/sec\] \[64=-32(0)+c\] \[64=c\] \[-32t+64=v(t)\] \[\int vdt=s(t)=\int (-32t+64) dt=-16t^2+64t+c\] \[s(0)=25\] \[25=-16(0)+64(0)+c\] \[25=c\] \[s(t)=-16t^2+64t+25\]
anonymous
  • anonymous
Please remember that calculus, one of the more predominant mathematics and the building block of much of the other fields was founded by a certain physicist that used it in much of his work =]

Looking for something else?

Not the answer you are looking for? Search for more explanations.