Conceptual Physics: Solving Problems in Math & Science

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The discussion revolves around solving conceptual physics problems, particularly focusing on motion, acceleration, and the effects of gravity. The user seeks guidance on how to approach problems involving acceleration, horizontal and vertical motion, and the application of relevant formulas. Key points include the importance of separating horizontal and vertical components when analyzing projectile motion and understanding that horizontal velocity remains constant while vertical motion is influenced by gravity. The user also expresses confusion about the relationships between velocity, acceleration, and time, as well as how to correctly apply formulas for different scenarios. Overall, the conversation emphasizes the need for a structured approach to problem-solving in physics.
  • #31
actually, the "d" should have been "-2", so the it should have been:

-2/-4.905=t*t
2/4.905=t²
 
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  • #32
is the d (-) because it is falling?
 
  • #33
Because you are starting from the shelf (y=0) to the ground (y=-2) and you use Vf-Vo for the difference.
 
  • #34
So now you have "t", put it in your X equation to get the distance from the edge of the shelf where it falls. And Voila !
 
  • #35
Physics is not that difficult you see :-)
 
  • #36
be back in 10 minutes, corner store time for some beers.
 
  • #37
ok, now i have

X
a=0
Vo=8m/s
V=8m/s
d=?
t=.64s

y

a=9.81
Vo=0
V=6.26
d=2m
t=.64s
 
  • #38
I keep trying to plug the equation only to cancel out most of the numbers and get D=T

When and if you get back on here, could you just run me through the steps when approached with these types of problems? maybe give me another example I could run through? I really appreciate the help here-
 
  • #39
d=Vo(or Vf it's the same) X t

remember, distance = velocity X time
 
  • #40
ok so my D=5.1 m/s/s
 
  • #41
You run a car at 100 mph for 2 hours. How much distance will you travel ?

d = 100 x 2 = 200 miles
 
  • #42
agreed, i understand the d=v/t formula, its the bigger one, that deals with this entire problem that i keep tripping on.
 
  • #43
d = 8 m/s X 0.64s

meters/seconds X seconds

seconds/seconds = nil

only meters stay in unit.
 
  • #44
yea, after i reread the problem i saw that i kept the s. So 5.1m is the distance the ball will end up at when i reaches the floor.
 
  • #45
Was it so difficult ? :-)
 
  • #46
totally! ahah, ok so i have a similar problem which has the same horiz velocity of 8m/s but instead height of the table they give me time(.24s). Do i work the problem the same way?
 
  • #47
or can i just use the D=v/t
 
  • #48
I think you have all the information you need.

Switch the variable in Y and resolve with the X equation like you did.
 
  • #49
d=8(.24)+1/2(o)(.24*.24)
 
  • #50
so the height of the table is 1.92m tall
 
  • #51
Hummm.. as I reread your question, it would be it be if I was the teacher I would ask for the height of the table.
 
  • #52
yea the problem is looking for the height of the shelf/table
 
  • #53
you again mixed the X and Y.

1.92m is where the ball will hit the ground from the shelf, horizontally.
 
  • #54
ok so the variables i have are

Vo=8m/s
a=0
t=.24s
d=1.92m

a=9.81m/s/s
 
  • #55
Gimme some minutes, I will draw something.
 
  • #56
I won't be leaving anytime soon, so take all the time you need. lol thanks.
 
  • #57
There.
 

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  • #58
See the big black line between X and Y ? That's on purpose. They are to be resolved SEPARATELY.
 
  • #59
k, i ll try out the problem, give me a sec or 2
 
  • #60
oops there's a error. it's "a = Vf X t² + Vo"

Sorry.

I was trying to make a good drawing lol
 

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