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

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- IsTim

Apparently I misinterpreted my question. So now I got a problem.

- TuringTest

is it 4.3

- IsTim

I took this as a regular acceleration question, and used \[V _{2}^{2}=v _{1^{2}}+2ad\]

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## More answers

- TuringTest

good...

- anonymous

how is it rolled off

- IsTim

Yeah, its 4.33.How?

- IsTim

No other info given. Previous questions tho.

- IsTim

Time is 0.396, and displacement is .763m.

- TuringTest

solve what you have for V2
V1 is zero
the number you get is the y-component of the velocity

- TuringTest

the x-component does not change

- IsTim

Im suppose to break it up into x and y components?

- IsTim

I got 5.799m/s (down)

- 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

rolls off*

- anonymous

reaction force?

- IsTim

How did you get such formulas? Both 2?

- 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

can angular rotation come into play?

- IsTim

I'm not that advd yet.

- anonymous

okies

- 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

@mth none of that stuff is necessary so why use it?

- anonymous

if air resistance negligable
gravity?

- IsTim

Gravity applies.

- TuringTest

gravity is in my formulation

- anonymous

on the rolling of the marble

- IsTim

So I rearrange V22=v12+2ad

- IsTim

Apparently I can't copy equations.

- anonymous

i was asking

- TuringTest

V22=v12+2ad
^^ ^^ ^ ^
Vy2 0 g h
leads to
Vy2=2gh

- anonymous

so potential energy does not play in part

- IsTim

im doing potential energy later. i hope.

- TuringTest

V22=v12+2ad
^^ ^^ ^ ^
Vy2 0 g h
*sorry got shifted around

- IsTim

So my result is 5.79m/s. What's my next course of action.

- TuringTest

where did you get that number?

- anonymous

sorry thats all i can do :(
my math sucks
i like theoretics much better

- IsTim

(1.93^3+2(9.8)(0.763))^1/2

- IsTim

IT's alright mth.

- IsTim

I re-did it, and got 4.32.

- 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

Oh. Shoot.

- 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

this time i got 3.87

- 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

Pythagorean Theorem. Got it.

- TuringTest

|dw:1327908694189:dw|Vy changes independently from Vx. That is a handy thing to know.
Good luck!

- IsTim

I got 4.32.

- TuringTest

4.326 ?

- IsTim

Ok. Thank yo uvery much.

- IsTim

4.3218

- IsTim

(3.867^2+1.93^2)^1/2

- TuringTest

\[\sqrt{1.93^2+2(9.8)(0.765)}\]is what I have in my calculator

- IsTim

Oops. Instead of .765, I used .763.

- TuringTest

Cheers :D

- IsTim

Yup. Thanks, if you ever decide to re-visit.

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