anonymous 3 years ago 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.

1. 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

2. anonymous

so its not $s=-16t ^{2}+v _{0}t+s _{0}$

3. anonymous

its same thing with

4. anonymous

with S0 as 25ft

5. anonymous

what does the $v _{0} s _{0}$ mean

6. anonymous

its v0*t here V0 is initaial velocity and here it is 64 ft/s

7. anonymous

so the equation would be S=-16t^2+64(1)+25

8. anonymous

yup but with t S=-16t^2+64(t)+25

9. 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

10. anonymous

yes

11. 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

12. 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

13. anonymous

Post such questions in physics group.. also refer to: http://openstudy.com/code-of-conduct

14. 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 $a=-32ft/sec$ $\int a dt=\int =-32dt=-32t+c=v(t)$ $v(0)=64 ft/sec$ $64=-32(0)+c$ $64=c$ $-32t+64=v(t)$ $\int vdt=s(t)=\int (-32t+64) dt=-16t^2+64t+c$ $s(0)=25$ $25=-16(0)+64(0)+c$ $25=c$ $s(t)=-16t^2+64t+25$

15. 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 =]