## 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

1. IsTim

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

2. TuringTest

is it 4.3

3. IsTim

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

4. TuringTest

good...

5. anonymous

how is it rolled off

6. IsTim

Yeah, its 4.33.How?

7. IsTim

No other info given. Previous questions tho.

8. IsTim

Time is 0.396, and displacement is .763m.

9. TuringTest

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

10. TuringTest

the x-component does not change

11. IsTim

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

12. IsTim

I got 5.799m/s (down)

13. 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$|V|=\sqrt{V_x^2+V_y^2}$

14. TuringTest

rolls off*

15. anonymous

reaction force?

16. IsTim

How did you get such formulas? Both 2?

17. 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

18. anonymous

can angular rotation come into play?

19. IsTim

I'm not that advd yet.

20. anonymous

okies

21. 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$\large v_y^2=v_{yi}^2+2gh=(0)^2+2gh\to v_y=\sqrt{2gh}$

22. TuringTest

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

23. anonymous

if air resistance negligable gravity?

24. IsTim

Gravity applies.

25. TuringTest

gravity is in my formulation

26. anonymous

on the rolling of the marble

27. IsTim

So I rearrange V22=v12+2ad

28. IsTim

Apparently I can't copy equations.

29. anonymous

30. TuringTest

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

31. anonymous

so potential energy does not play in part

32. IsTim

im doing potential energy later. i hope.

33. TuringTest

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

34. IsTim

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

35. TuringTest

where did you get that number?

36. anonymous

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

37. IsTim

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

38. IsTim

IT's alright mth.

39. IsTim

I re-did it, and got 4.32.

40. 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.

41. IsTim

Oh. Shoot.

42. 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.

43. IsTim

this time i got 3.87

44. 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...

45. IsTim

Pythagorean Theorem. Got it.

46. TuringTest

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

47. IsTim

I got 4.32.

48. TuringTest

4.326 ?

49. IsTim

Ok. Thank yo uvery much.

50. IsTim

4.3218

51. IsTim

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

52. TuringTest

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

53. IsTim

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

54. TuringTest

Cheers :D

55. IsTim

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