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IsTim
 4 years ago
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
IsTim
 4 years ago
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

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IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Apparently I misinterpreted my question. So now I got a problem.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I took this as a regular acceleration question, and used \[V _{2}^{2}=v _{1^{2}}+2ad\]

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0No other info given. Previous questions tho.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Time is 0.396, and displacement is .763m.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2solve what you have for V2 V1 is zero the number you get is the ycomponent of the velocity

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2the xcomponent does not change

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Im suppose to break it up into x and y components?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2they 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\[V=\sqrt{V_x^2+V_y^2}\]

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0How did you get such formulas? Both 2?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2like I said, the horizontal component is constant because there is no force acting on the object in the xdirection, so the horizontal velocity is the 1.93m/s it rolled of the table with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can angular rotation come into play?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2the ycomponent is what your formula leads to for the updown 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
 4 years ago
Best ResponseYou've already chosen the best response.2@mth none of that stuff is necessary so why use it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if air resistance negligable gravity?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2gravity is in my formulation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on the rolling of the marble

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0So I rearrange V22=v12+2ad

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Apparently I can't copy equations.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h leads to Vy2=2gh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so potential energy does not play in part

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0im doing potential energy later. i hope.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2V22=v12+2ad ^^ ^^ ^ ^ Vy2 0 g h *sorry got shifted around

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0So my result is 5.79m/s. What's my next course of action.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2where did you get that number?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry thats all i can do :( my math sucks i like theoretics much better

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0(1.93^3+2(9.8)(0.763))^1/2

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0I redid it, and got 4.32.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I 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.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2if 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.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2\[V_x=1.93\]\[V_y=\sqrt{2gh}\]\[V=\sqrt{V_x^2+V_y^2}=\sqrt{1.93^2+2(9.8)(0.765)}\approx4.33\]I'm going to sleep here so I gotta wrap it up...

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Pythagorean Theorem. Got it.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1327908694189:dwVy changes independently from Vx. That is a handy thing to know. Good luck!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{1.93^2+2(9.8)(0.765)}\]is what I have in my calculator

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Oops. Instead of .765, I used .763.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Yup. Thanks, if you ever decide to revisit.
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