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use the equations of motion with constant accleeration
s = ut + 0.5gt^2
what are the equations if you dont mind:)
s = displacement = 0 g = -32 ft/s62 and u = 25 ft/s
0 = 25t - 16t^2 solve for t
cwrw238 you are so wrong that newton cam back to life and died again
so what would the answer be
Michelle i would like to help you but i think that the data is incomplete...
but that is how the question is written out what am suppose to do? i could go back and reread the lesson
yes you should you dont even know what equations of motion are
you see this is my third time taking this class and i really don't want to fail again.
Ahaha michelle take it easy!
i think I will, look over the lesson again,but what do you guys think would be the best way to approach it.
The problem that i am having is the fact that they didnt mention at which point in height the balls stops and starts falling back
yeah, i guess, i could always email my teacher about it because there are alway's glitches like that happening in the tests.
cwrw238 already said it solve 0 = 25t - 16t^2 for t to get your answer (it will be the root that is not 0)
um ok, i will go back and do it that way.
thanks for the advice everyone, you all have been great so you have a nice day
YOU HAVE TO USE SEQENCES EVERTHING ELSE IS WRONG!!!!!!!!!!!!!!
16t^2 - 25 = 0 t(16t - 25) = 0 16t = 25 f = 25/16 = 1.6