IsTim
  • IsTim
A marble rolls off a table with a velocity of 1.93m/s[horizontally]. The tabletop is 76.5cm above the floor. If air resistance is negligible, determine the velocity at impact
Physics
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SOLVED
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katieb
  • katieb
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IsTim
  • IsTim
Apparently I misinterpreted my question. So now I got a problem.
TuringTest
  • TuringTest
is it 4.3
IsTim
  • IsTim
I took this as a regular acceleration question, and used \[V _{2}^{2}=v _{1^{2}}+2ad\]

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TuringTest
  • TuringTest
good...
anonymous
  • anonymous
how is it rolled off
IsTim
  • IsTim
Yeah, its 4.33.How?
IsTim
  • IsTim
No other info given. Previous questions tho.
IsTim
  • IsTim
Time is 0.396, and displacement is .763m.
TuringTest
  • TuringTest
solve what you have for V2 V1 is zero the number you get is the y-component of the velocity
TuringTest
  • TuringTest
the x-component does not change
IsTim
  • IsTim
Im suppose to break it up into x and y components?
IsTim
  • IsTim
I got 5.799m/s (down)
TuringTest
  • TuringTest
they are already broken up\[V_x=1.93\]because it rolls of horizontally, and that component is constant because no force acts in that direction. The vertical component is\[V_y=\sqrt{2gh}\]and the total velocities magnitude is\[|V|=\sqrt{V_x^2+V_y^2}\]
TuringTest
  • TuringTest
rolls off*
anonymous
  • anonymous
reaction force?
IsTim
  • IsTim
How did you get such formulas? Both 2?
TuringTest
  • TuringTest
like I said, the horizontal component is constant because there is no force acting on the object in the x-direction, so the horizontal velocity is the 1.93m/s it rolled of the table with
anonymous
  • anonymous
can angular rotation come into play?
IsTim
  • IsTim
I'm not that advd yet.
anonymous
  • anonymous
okies
TuringTest
  • TuringTest
the y-component is what your formula leads to for the up-down portion of the motion if you recognize that the initial downward velocity is zero\[\large v_y^2=v_{yi}^2+2gh=(0)^2+2gh\to v_y=\sqrt{2gh}\]
TuringTest
  • TuringTest
@mth none of that stuff is necessary so why use it?
anonymous
  • anonymous
if air resistance negligable gravity?
IsTim
  • IsTim
Gravity applies.
TuringTest
  • TuringTest
gravity is in my formulation
anonymous
  • anonymous
on the rolling of the marble
IsTim
  • IsTim
So I rearrange V22=v12+2ad
IsTim
  • IsTim
Apparently I can't copy equations.
anonymous
  • anonymous
i was asking
TuringTest
  • TuringTest
V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h leads to Vy2=2gh
anonymous
  • anonymous
so potential energy does not play in part
IsTim
  • IsTim
im doing potential energy later. i hope.
TuringTest
  • TuringTest
V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h *sorry got shifted around
IsTim
  • IsTim
So my result is 5.79m/s. What's my next course of action.
TuringTest
  • TuringTest
where did you get that number?
anonymous
  • anonymous
sorry thats all i can do :( my math sucks i like theoretics much better
IsTim
  • IsTim
(1.93^3+2(9.8)(0.763))^1/2
IsTim
  • IsTim
IT's alright mth.
IsTim
  • IsTim
I re-did it, and got 4.32.
TuringTest
  • TuringTest
I keep trying to tell you that the equations are separate Vx=1.93 Vy=(2*9.890.763)^1/2 you have Vx plugged in where the initial speed goes, which is zero.
IsTim
  • IsTim
Oh. Shoot.
TuringTest
  • TuringTest
if you wanted both those things in the same equation it would have to be a vector formula, and I don't think you've done parametric stuff yet.
IsTim
  • IsTim
this time i got 3.87
TuringTest
  • TuringTest
\[V_x=1.93\]\[V_y=\sqrt{2gh}\]\[|V|=\sqrt{V_x^2+V_y^2}=\sqrt{1.93^2+2(9.8)(0.765)}\approx4.33\]I'm going to sleep here so I gotta wrap it up...
IsTim
  • IsTim
Pythagorean Theorem. Got it.
TuringTest
  • TuringTest
|dw:1327908694189:dw|Vy changes independently from Vx. That is a handy thing to know. Good luck!
IsTim
  • IsTim
I got 4.32.
TuringTest
  • TuringTest
4.326 ?
IsTim
  • IsTim
Ok. Thank yo uvery much.
IsTim
  • IsTim
4.3218
IsTim
  • IsTim
(3.867^2+1.93^2)^1/2
TuringTest
  • TuringTest
\[\sqrt{1.93^2+2(9.8)(0.765)}\]is what I have in my calculator
IsTim
  • IsTim
Oops. Instead of .765, I used .763.
TuringTest
  • TuringTest
Cheers :D
IsTim
  • IsTim
Yup. Thanks, if you ever decide to re-visit.

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