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nburgher
Group Title
HELP PLEASE!!!!!!!!!! If a rock is thrown upward with an initial velocity of 64 feet per second from the top of a 25foot building, write the height (s) equation using this information.
 one year ago
 one year ago
nburgher Group Title
HELP PLEASE!!!!!!!!!! If a rock is thrown upward with an initial velocity of 64 feet per second from the top of a 25foot building, write the height (s) equation using this information.
 one year ago
 one year ago

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matricked Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

nburgher Group TitleBest ResponseYou've already chosen the best response.0
so its not \[s=16t ^{2}+v _{0}t+s _{0}\]
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
its same thing with
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
with S0 as 25ft
 one year ago

nburgher Group TitleBest ResponseYou've already chosen the best response.0
what does the \[v _{0} s _{0} \] mean
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
its v0*t here V0 is initaial velocity and here it is 64 ft/s
 one year ago

nburgher Group TitleBest ResponseYou've already chosen the best response.0
so the equation would be S=16t^2+64(1)+25
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
yup but with t S=16t^2+64(t)+25
 one year ago

nburgher Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

nikhilchabe Group TitleBest ResponseYou've already chosen the best response.0
Post such questions in physics group.. also refer to: http://openstudy.com/codeofconduct
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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 \[a=32ft/sec\] \[\int a dt=\int =32dt=32t+c=v(t) \] \[v(0)=64 ft/sec\] \[64=32(0)+c\] \[64=c\] \[32t+64=v(t)\] \[\int vdt=s(t)=\int (32t+64) dt=16t^2+64t+c\] \[s(0)=25\] \[25=16(0)+64(0)+c\] \[25=c\] \[s(t)=16t^2+64t+25\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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 =]
 one year ago
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