## idku one year ago Ok, can someone show me ....

1. idku

|dw:1441289020025:dw|

2. idku

I found: Horizontal component of the velocity: $$\LARGE V_x=11\cos(44)\approx 7.9$$ Vertical component of the velocity: $$\LARGE V_x=11\sin(44)\approx 7.6$$ ---------------------------------------------------- How do I find the maximum height that the bsll goes above the ground?

3. idku

bsll is supposed to say ball.

4. ganeshie8

What's the question ?

5. idku

How do I find the maximum height that the bAll goes above the ground?

6. anonymous

Do you know your equations of motion?

7. idku

Which one?

8. idku

I might as well just say, no I don't....

9. ganeshie8

i think we need to know the height of those ppl

10. ganeshie8

at least we need to know the height of the person throwing the ball

11. anonymous

This is projectile motion correct, so it's better if you label your vector quantities more precisely, so we're dealing with the y - direction here$\vec a = \frac{ v_{yf}-v_{yi} }{ t }$ where $v_{yf}=0$ is there more information to this, all you gave is a diagram?

12. idku

f is final, i is initial. just clarifying....

13. idku

I deleted the question I am so bump ---------------------------- Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

14. idku

Wait, a is acceleration, but for maximum height, why do we need it? (btw, sorry again and again for deleted the question.)

15. ganeshie8

The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height

16. ganeshie8

$v_y = u_y - gt$ set that equal to 0 and find the time at which that happens

17. anonymous

a = g in this case

18. idku

what is u_y ?

19. ganeshie8

$$a = -g$$ because we take upwards as increasing position

20. ganeshie8

$$u_y$$ = initial vertical component of velocity

21. idku

oh. my teacher labelled it $$\LARGE y_{0_{^{_{\huge _x}}}}$$

22. anonymous

You can label it what ever, long as it makes sense to you and your teacher

23. idku

or, idk maybe I am confusing labels. I suck at this:)

24. ganeshie8

thats ridankulous normally we use $$v$$ and $$u$$ for velocity $$x$$ and $$y$$ for positions

25. idku

Yeah it is I that mixed something up (as always)

26. anonymous

we never used u, always v_0 xD

27. idku

Oh yes V_(0y), that!

28. ganeshie8

v = u + at is how i still remember my kinematics eqns from highschool

29. anonymous

that's much easier to

30. ganeshie8

it doesn't matter but we all need to agree on same thing..

31. idku

$$\LARGE V_y=V_{0y}-gt$$ $$\LARGE 0=V_{0y}+9.81t$$ and what did you say set equal to 0? like this, or am I wrong again?

32. ganeshie8

looks good and do you know why you're setting that equal to 0

33. ganeshie8

no wait, you flipped the sign

34. ganeshie8

$$\LARGE V_y=V_{0y}-gt$$ $$\LARGE 0=V_{0y}\color{red}{-}9.81t$$ and what did you say set equal to 0? like this, or am I wrong again?

35. idku

(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)

36. idku

oh, you said -g, so I thought... ok will take a note

37. idku

$$\LARGE 0=V_{0y}+9.81t$$

38. idku

And I am setting it to 0, because V_y is the derivative of D_y (where D is displacement - in ase of y, the height)

39. idku

So we are already having the derivative of the function that we want to maximize. Right?

40. idku

41. idku

$$\LARGE 0=V_{0y}-9.81t$$

42. ganeshie8

remember this : $$g$$ is acceleration due to gravity, it equals approximately $$9.81$$ on surface on earth, it is always positive.

43. idku

wait, so no minus; plus?

44. ganeshie8

$$a$$ is acceleration, in our case $$a = -g = -9.81$$

45. idku

$$\Large 0=V_{0y}\color{red}{\bf-}9.81t$$

46. idku

so g is pos, -g is neg. at least I know that. I am a pes(t) - imist.

47. ganeshie8

$$a$$ is negative because $$g$$ is pointing downward but we are taking "upwards" as increasing position

48. idku

Ok, and then I just find cthe maximum of our unknown displacement of y, based on the given velocity component (which is the derivative of the displacement)

49. ganeshie8

right, one thing at a time first find the time at which vertical component of velocity becomes 0

50. idku

t=0?

51. idku

at t=0 ... V_(0y) is 0.

52. ganeshie8

really ? at $$t=0$$ the ball surely has nonzero vertical component of velocity : |dw:1441290728442:dw|

53. idku

Wait, what is my equation for this? (to answer your question about the t-value at which the vertical velocity is equivalent to zero.)

54. ganeshie8

Easy, as you said, vertical component of velocity of ball is modeled by the function : $V_y=V_{0y}-9.81t$ you're "given" $$V_{0y} = 11\sin(44)$$, therefore the function becomes $V_y=11\sin(44)-9.81t$

55. idku

oh, can I try?

56. ganeshie8

let me finish typing first

57. idku

58. ganeshie8

il let u work it but im not finished yet and i think u need to get some clarifications :)

59. ganeshie8

you know that $$V_y$$ is $$0$$ when the ball reaches maximum height so use that fact to find the "time" at which the ball reaches maximum height

60. ganeshie8

once you have the "time" at which ball reaches the maximum height, you can use that to find the required maximum height

61. anonymous

This is a nice way to work with vertical trajectory you could also still do it the way I was showing above, the main thing is knowing the final velocity of the y component = 0 at maximum height

62. ganeshie8

that is all i wanted to say go ahead, set $$V_y$$ equal to $$0$$ and solve $$t$$

63. idku

$$\LARGE V_y=11\sin(44)-9.81t$$ $$\LARGE 0=11\sin(44)-9.81t$$ $$\LARGE 11\sin(44)=9.81t$$ $$\LARGE t=\frac{9.81}{11\sin(44)}$$

64. idku

hope at least that I have managed:)

65. anonymous

Hmm

66. idku

oh no

67. idku

well that to the power of -1

68. anonymous

It should be $t = \frac{ 11\sin(44) }{ 9.81 }$

69. idku

$$\LARGE t=\frac{11\sin(44)}{9.81}$$ I am even dumber than what I thought

70. idku

yes, I am dividing by 9.81 don't know what's up with my Brain

71. idku

$$\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78$$

72. anonymous

You could actually go ahead and use $y = v_{0y} t$ now

73. anonymous

That will give you your maximum height

74. idku

$$\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96$$

75. idku

calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.

76. idku

That is the max height....

77. idku

Let me make an attempt to recap what I have done for maximum height ((My suffering is not equivalent of any sin (of $$\theta$$), I hate this!)) ----------------------------------------------------- RECAP, MAX HEIGHT $$\LARGE V_y=V_{0y}-gt$$ $$\LARGE 0=11\sin(44)-9.81t$$ $$\LARGE t =0.78$$ so max height occrs at t=0, and now we plug that into vertical displacement $$\LARGE y=V_{0t}\times t$$ $$\LARGE y=11\sin(44)\times 0.78=5.96$$

78. anonymous

t=0?

79. idku

sorry, what do you mean?

80. anonymous

Why did you write max height is at t = 0

81. idku

oh, occurs at t=0.78

82. anonymous

Don't round until the final answer

83. anonymous

So use the full time rather than 0.78

84. idku

$$\LARGE y=\sin^2(44)/9.81$$ ?

85. idku

I forgot 11

86. idku

0.78 was really 11sin(44)/9.81 So y is y=11sin(44)×11sin(44)÷9.81

87. idku

well, if that is right then it is still 5.95

88. idku

$$\Large y = v_{0y} t - \frac{ 1 }{ 2 } g t^2$$ $$\Large y = 11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2$$ http://www.wolframalpha.com/input/?i=11sin%2844%29%C3%97%2811sin%2844%29%2F9.81%29-%281%2F2%29%289.81%29%2811sin%2844%29%2F9.81%29%C2%B2 I get 2.97597 ----> 2.98

89. idku

And that is not correct either -:(

90. anonymous

Hold on, let just use $h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) }$ we also have to remember the 1.5m forgot about that

91. idku

I don't think that there is another explanation for an incorrect output from the formula, other than that I failed to find all the values correctly. I don't think formulas are faulty. (I start to despair from thinking that I can think ) HIHPHh9hduhoduhdue89wv9vi

92. idku

I have to go, I will come back about 2 h later-;(

93. anonymous

Yes, this should work, the problem was we did not take account of the question, you must read it carefully, and not make assumptions. Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. $h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) } = \frac{ [11\sin(44)]^2 }{ 2(9.81)} = 2.975...$ but we must add the 1.5m as well to this, so we get 2.975...+1.5 = 4.4759m = 4.5m

94. anonymous

I think you should use this though as we found time for this precisely $h = \frac{ 1 }{ 2 }g t^2 = \frac{ 1 }{ 2 }(9.81)(0.78)^2 = 2.965....+1.5 = 4.47 = 4.5m$ so it works!

95. ganeshie8

below equation for vertical position perfectly fine $$\Large y = \color{red}{y_0}+v_{0y} t - \frac{ 1 }{ 2 } g t^2$$ $$\Large y = \color{red}{y_0}+11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2$$

96. anonymous

When you put your final answer, make sure it's 2 sig digs!

97. idku

Oh, if it was something on a same level of thinking then I would have probably figured. Yes the 1.5 meters of the ground, because my origin is (0, 1.5) Iambatman tnx! I am checking it and it is.....