In need help with this question. All the other answers didn't work!?
824568
We can use trigonometry to determine the vertical and horizontal components of the stone's velocity.  This will let us determine how far it goes.\n\nImagine a right triangle with a hypotenuse of 14 units and one angle of 26 degrees.  The side opposite the 26 degree angle will be as long as the stone's vertical velocity.  The side adjacent to the 26 degree angle will be as long as the stone's horizontal velocity.\n\nSince:\n\nsinA = opposite / hypotenuse... \nsin26 = vertical velocity / 14...\nvertical velocity = 6.14m/s\n\ncosA = adjacent / hypotenuse;\ncos26 = horizontal velocity / 14...\nhorizontal velocity = 12.56m/s\n\nIf the object had an initial vertical velocity of 6.14m/s, we know how long it stayed in the air because:\nacceleration = change in velocity / time\nWhich is the same as saying:\ntime = change in velocity / acceleration\nThis applies in any situation where the acceleration is constant.\nSo, if we want to know how long it took for the stone to reach 0m/s vertical velocity (the point at which it stops rising and starts falling), the change in velocity will be 6.14m/s.  If we just divide this number by acceleration due to Earth's gravity (9.81m/s^2), we get 0.63 seconds.  The stone traveled upward for 0.63 seconds.  This means that it must have also traveled downward for 0.63 seconds before hitting the ground, for a total flight time of 1.26 seconds.\n\nNow, we have a simple velocity equation.  Since we know that:\ndistance = velocity * time \nSince we're only worried about how far the stone traveled horizontally, we use the stone's horizontal velocity: 12.56m/s.  Time will be the total time that the stone was moving: 1.26sec.\n\nSo...\ndistance = 12.56 * 1.26 = 15.83 meters.\n\nHope that helped!
191038
hmm.. sounds like a homework question.  tsk tsk..
185330
I think you might want to rethink your question and restate it. I can't understand it, and it is unlikely that anyone else can, either.\n\nGood luck!
693416
i recommend to just seach it by Yahoo! search than just asking it here in Yahoo answeres... it would take time to type all what you are asking...
32595
Sorry, but you should really do your homework by yourself.  Try using google.
1012727
I found some resources that'll assist you further -- see the source box.\n\nHope that helps!
870037
Sorry, I have tried to read this to find an actual question and can't. Please try adding details, use standard English and punctuation and the spell-checker and maybe someone can help you.
567825
Go to this web page I think it might give you all your answers:\n\nhttp://library.thinkquest.org/19539/randj.htm
945377
Looks like you had too many words for Yahoo! Answers.  Next time try using the additional details.
1030000
Sounds like homework hehe... this is too long to answer in full.
53838
I'll be glad to help. Could you provide more detail? Do you remember any examples of the (something ridiculous) that I could work with? Thanks. If you can get me more details I'll edit my answer if I find something.
1050777
Here is a start-----\nYou need to do your own homework.
1186158
look in a web site..if u want help.\nidont know.
457702
I think you need to start doing your own homework because I've checked through your question log and you've posted a question like this three times.  Get off your butt and do your own work.
781038
I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
196338
This is such a run-on question!!!!!!  Could you please re-phrase it?
1088601
You should have posted this in the Homework Help section. \n\nHowever, you shouldn't post the complete question anyway: no-ones going to do your homework for you, only help you out!