## idku one year ago Math - Based - Physics question.

1. idku

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2. idku

NOTE: in the corner of the picture 2, all it says is "1.5 m".

3. IrishBoy123

well you now know that its horizontal velocity is 9m/s and that it travels a vertical distance of 5.5m before it stops rising that second piece of info allows you to calculate its vertical velocity as it leaves her hand use the equation of motion $$v^2 = u^2 + 2ax$$ v = 0, a = -g, x = 5.5m, so get u

4. IrishBoy123

you follow? or questions?

5. idku

First of, how do I know that it travles 5.5 meters before it stops rising?

6. IrishBoy123

the question says it rises to 7m above ground, it start from 1.5m above ground, 7 - 1.5 = 5.5m

7. idku

Oh, ok, 5.5m is our vertical displacement. Good. And what does "u" stand for?

8. IrishBoy123

u = initial velocity v = final velocity yet another notation but it's simple and really easy on the eye eg because no subscripts

9. idku

Oh, good. And v=0 because that is the point where it changes from positive (the height of the ball increasing) to negative (the height of the ball decreasing). Am I right?

10. IrishBoy123

yes

11. idku

$$\large v^2=u^2+2ax$$ $$\large 0^2=u^2+2(-9.81)(5.5)$$ like this?

12. idku

or, am I plugging +9.81 ?

13. IrishBoy123

you got it right!

14. idku

the first time?

15. IrishBoy123

this is right $\large 0^2=u^2+2(-9.81)(5.5)$

16. idku

$$\large v^2=u^2+2ax$$ $$\large 0^2=u^2+2(-9.81)(5.5)$$ $$\large 0^2=u^2-107.91$$ $$\large u=\sqrt{107.91}$$

17. idku

yes, if I plugged in +9.81 I would have got imaginary velocity.

18. idku

u=10.39 m/s (this is initial velocy)

19. IrishBoy123

*yes, if I plugged in +9.81 I would have got imaginary velocity.* nice! i'm not doing the sums so i trust you are happy with your calculation?

20. IrishBoy123

OK, really important now. right now, we are working in the vertical direction! yes?!

21. IrishBoy123

so "u=10.39 m/s (this is initial VERTICAL velocy)" [assuming your calculator is working!]

22. idku

I used wolfram, so should work, but the practice thingy doesn't accept 10.39

23. IrishBoy123

we already know the HORIZONTAL velocity, which is constant yes?

24. IrishBoy123

we have to combine vertical and horizontal using pythagoreas

25. idku

I am not given the angle, how would I tell horizontal velocty/

26. idku

?

27. idku

Sorry.

28. IrishBoy123

i told you it in my first post go take a look and see where it came from

29. IrishBoy123

in case you're not getting this i will repeat it:p we already know the HORIZONTAL velocity, which is constant yes? i told you it in my first post go take a look and see where it came from bandwidth is an issue for me :-(

30. idku

I got disconnected due to lightning. My apologies, I am here....

31. idku

yes, I see so I got 13.75 m/s http://www.wolframalpha.com/input/?i=%E2%88%9A%289%C2%B2%2B10.39%C2%B2%29

32. IrishBoy123

no worries! hope everyone's OK.

33. idku

No one is outside:) My garage is inside the house, so it's alright.

34. idku

I wish I had you as my physics teacher, you made everything as easy as even a non-math person such as myself got it!

35. idku

There is just one more question (for this part at least), if you don't mind: How high above the ground will the ball be when it gets to Julie?

36. IrishBoy123

excellent, one thing you **must** remember we do the horizontal axis and vertical differently no acceleration in horizontal, g acts in the vertical then combine through pythagoreas that applies throughout mechanics and kinematics...

37. IrishBoy123

well you know how long it will take to get to Julie, right

38. idku

ok, i will take a note of pythagorean as regards to velocities as well as distances....

39. IrishBoy123

brill! you know ho far apart they are? you know the horizontal velocity? and so?

40. idku

Horizontal velocity is 9m/s. And the distance between them would be?

41. IrishBoy123

didn't you work that out in an earlier part of the question?

42. IrishBoy123

use that number

43. idku

I did the maximum height for [the part in] picture 1, and I did horizontal and vertical displacements for that (and this) part(s).

44. idku

wait, is it 11cos(44) ?

45. idku

but that is the velocity of the first through, not the distance. So that is not it.

46. idku

9(the horiz. velocity of the first through))

47. IrishBoy123

let me work it out first, it might jog your memory just wait

48. IrishBoy123

OK from the very first question, we know that it was thrown at 11m/s at 44deg we can use equation ofmotion to get flight time $x = ut + \frac{1}{2}a t^2 = t(u + \frac{1}{2}at)$ follow so far? we are working in the vertical direction now.

49. idku

oh, so the displacement-for-y formula and displacement-for-x formula are all same, besides that every y is now x instead?

50. IrishBoy123

always in these x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time always here we are tracking the ball from the first girl to the second, and we are looking at motion in the vertical direction, where gravity plays a role

51. idku

wait, you said: $$x = ut + \frac{1}{2}at^2$$ so x in this case is the vertical distance I am solving for. a = -g = -9.81m/s u = 9m/s  t = ? how do I figure out time?

52. IrishBoy123

we need to back to the first question to get the distance between the girls, because we know that one threw it and one caught it. so we know it started t 1.5m above ground, and ended there when caught. we also know it was thrown at 11m/s at 44drg to horizontal.

53. IrishBoy123

and to do that we need first to calculate how long the ball spent in the air travelling between the girls there's a lot in there so take your time

54. idku

While the distance is same, the times are different at different speeds. So we can't use the first question to calculate the time of second through. (That would have been rediculous if I have done so).

55. idku

But, distance I can use that same displacement formula, applying everything I know from picture 1?

56. IrishBoy123

no we need to know how far the girls are apart we cannot do that from the information in the second part of the question we need to go right back to the first part to do that that is why i suspect that you have already calculated that distance

57. idku

distance = 12.32

58. idku

meters*

59. idku

$$12.32 = 9t + \frac{1}{2}(-9.81)t^2$$ ?

60. idku

and olve for time?

61. idku

solve

62. IrishBoy123
63. IrishBoy123

OK so back to this last bit and we need to recap, i think. we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s yes now how long will it take to reach the first girl? in the horizontal direction....

64. idku

Oh, I am setting incorrect equation, because we are working the time based on the information we have from the first question, so u=11

65. idku

and I said u=9

66. idku

$$12.32 = 11t + \frac{1}{2}(-9.81)t^2$$ But I still get imaginary solutions

67. idku

with +9.81 I get real solutions:)

68. IrishBoy123

just stop for a minute, i think we are crossing posts:p we know the distance between them is 12.327m we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s agreed?

69. idku

yes, we had all of this either as given or found that:)

70. IrishBoy123

so now we need to know how long that will take, for a ball with horizontal velocity = 9m/s to travel the distance between the girls, which is 12.327m

71. idku

$$12.327= 9t + \frac{1}{2}(-9.81)t^2$$

72. IrishBoy123

there is no gravity in the horizontal direction:p

73. IrishBoy123

just constant velocity in horizontal direction

74. idku

what is my formula for x then?

75. idku

which formula do I use, I mean

76. IrishBoy123

oooHHHHH!!!! think!

77. idku

$$x = ut + \frac{1}{2}t^2$$ ?

78. IrishBoy123

the first one you learn!

79. idku

Distance = Velocity × Time ?

80. IrishBoy123

$x = ut + \frac{1}{2}(0)t^2 = ut$ or t = distance / velocity !!!!

81. IrishBoy123

yeah!!

82. idku

oh, for horizontal veloccity the acceleration is 0. distance = (initial velocity) × (time) ---> time = distance ÷ (initial velocity) time = 12.327m ÷ 9m/s (when dividing by m/s, you multiply times s/m, so meters cancel we get: (12.327/9)s )

83. idku

s is seconds.

84. idku

for whatever that quotient is

85. IrishBoy123
86. idku

≈ 1.37

87. idku

yes:)

88. idku

Now we need to apply t=1.37 (time length interval of the ball's second through) to find the height of the ball when it gets to julie - that is when it travels 12.327 meters.

89. IrishBoy123

one final recap before we finish we know the distance between them is 12.327m [but we don't need that info anymore] we know that the second girl threw it with -horizontal velocity = 9m/s - vertical velocity = 10.39 m/s we know that the journey in th e**horizontal** direction too 1.37s. we need to know what happens in the vertical direction at t = 1.37. at what height is the ball? agreed?

90. idku

yes, I undertand that! You are a wonderful person!

91. IrishBoy123

so we need an equation of motion with acceleration in it because of gravity $x = ut + \frac{1}{2}at^2$

92. IrishBoy123

we want to know x. x is the displacement of the ball in the vertical direction a = -9.91 m/s/s t = 1.37 s u = 10.39 m/s

93. idku

x is the variable of interest. initial velocity -- u=10.39m/s (we are interested in vertical velocity) time -- t=1.37s (time is same for Vert. and Horiz. displacement of the ball) acceleration [due to gravity] -- a=-(g)=-(9.81)m/s² $$x = 10.39m/s \times 1.37s + \frac{1}{2}(-9.81m/s^2)(1.37s)^2$$

94. idku

so in the first and second part we have meters left only. Just making sure t makes sense the way I do it, so that if I encounter unit confusion I know I have done something wrong. $$x = 10.39 \times 1.37 + \frac{1}{2}(-9.81)(1.37)^2$$

95. idku

and whatever that is equal to....

96. idku

5.028m

97. idku
98. idku

+1.5m

99. IrishBoy123

yes so she throws it 5.03m above the first girls hand. add 1.5m onto that if you want distance above ground

100. idku

I get 5.028+1.5=6.528

101. IrishBoy123

oh! good you got it yourself:p

102. IrishBoy123

well, that's the right "way" to do it, any mistakes are calculations.....we now know "how" to do it:p

103. idku

so the main things I need are: $$\large v^2=u^2+2ax$$ $$\large x=ut+\dfrac{1}{2}at^2$$

104. idku

and of course I know how to find horizontal and vertical velocity components based on a given angle and initial velocity....

105. IrishBoy123

let me type them out, there are 3 $v = u + at$ $x = ut + \frac{1}{2}at^2$ $v^2 = u^2 + 2ax$ trick is to pick the right one to use as they all look at different things. eg the last one doesn't have t in it. but it is incredibly useful if you know of it.

106. idku

this is probably the most valuable post in my life. TY so much. No excuse me I got to go:) tnx for formula recap!

107. IrishBoy123

yes - horizotal vs vertical there's an awful awful lot of info in there, you have done well! cheers!!

108. idku

$$v = u + at$$ $$v_x~~ =~~u~ \cos(\theta)$$ $$v_y ~~=~~ u~\sin(\theta)$$ $$d = ut + \frac{1}{2}at^2$$ $$v^2 = u^2 + 2ax$$ x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time Ok, when I have to time to look more into it, i will. Thank you very much. I actually get this stuff now.