Question

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

Answer 1

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

Answer 2

is it 4.3

Answer 3

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

Answer 4

good...

Answer 5

how is it rolled off

Answer 6

Yeah, its 4.33.How?

Answer 7

No other info given. Previous questions tho.

Answer 8

Time is 0.396, and displacement is .763m.

Answer 9

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

Answer 10

the x-component does not change

Answer 11

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

Answer 12

I got 5.799m/s (down)

Answer 13

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\[|V|=\sqrt{V_x^2+V_y^2}\]

Answer 14

rolls off*

Answer 15

reaction force?

Answer 16

How did you get such formulas? Both 2?

Answer 17

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

Answer 18

can angular rotation come into play?

Answer 19

I'm not that advd yet.

Answer 20

okies

Answer 21

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\[\large v_y^2=v_{yi}^2+2gh=(0)^2+2gh\to v_y=\sqrt{2gh}\]

Answer 22

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

Answer 23

if air resistance negligable
gravity?

Answer 24

Gravity applies.

Answer 25

gravity is in my formulation

Answer 26

on the rolling of the marble

Answer 27

So I rearrange V22=v12+2ad

Answer 28

Apparently I can't copy equations.

Answer 29

i was asking

Answer 30

V22=v12+2ad
^^ ^^ ^ ^
Vy2 0 g h

leads to
Vy2=2gh

Answer 31

so potential energy does not play in part

Answer 32

im doing potential energy later. i hope.

Answer 33

V22=v12+2ad
^^ ^^ ^ ^
Vy2 0 g h

*sorry got shifted around

Answer 34

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

Answer 35

where did you get that number?

Answer 36

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

Answer 37

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

Answer 38

IT's alright mth.

Answer 39

I re-did it, and got 4.32.

Answer 40

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.

Answer 41

Oh. Shoot.

Answer 42

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.

Answer 43

this time i got 3.87

Answer 44

\[V_x=1.93\]\[V_y=\sqrt{2gh}\]\[|V|=\sqrt{V_x^2+V_y^2}=\sqrt{1.93^2+2(9.8)(0.765)}\approx4.33\]I'm going to sleep here so I gotta wrap it up...

Answer 45

Pythagorean Theorem. Got it.

Answer 46

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