## YumYum247 one year ago elp please!!!!

1. YumYum247

Can someone please check question number 41 for me plz?!?!?!!?!?

2. YumYum247

@dan815

3. dan815

can i suggest one thing

4. dan815

it might look like its less work with the formulas and stuff, but there is a simpler way to think about these questions, even though the work might seem like its a bit more

5. YumYum247

i don't care, spit it out!!!!

6. dan815

okay look

7. YumYum247

i am.....^_^

8. dan815

|dw:1439951841864:dw|

9. YumYum247

ive already done that....how do i figureout the horizontal displacement of the projectile.....??? :)

10. dan815

now u know the horizontal speed Vx, so what we need to do is find the time it is in the air for, and multiply by that time

11. YumYum247

yes those are the initial vertical velocity and initial horizontal velocity.....

12. dan815

and to find how long its in Air for, we know the Vertival speed and we have the accerelation

13. dan815

one thing about the launch is that the starting and ending speed for the vertical is the same

14. dan815

jstu the direction is opposite

15. YumYum247

yup i know that.... :)

16. dan815

so what we need to do is see the time it takes for the velicity to go to 0 and mulitply by 2

17. dan815

our acc is 9.8 m/s/s so Vy= 35*sin 40 lets see how long that takes to go to 0 34*sin 40 - 9.8*t = 0 every second we are going to lose 9.8 see how many secs that will take find that time multiply by 2

18. dan815

35*sin 40 - 9.8*t = 0 t=35sin(40)/9.8 it will be up in air twice that time so our horizontal distanve must be ? Vx*2t = 35*cos(40) * 2*35sin(40)/9.8

19. dan815

comesout to 123.101

20. YumYum247

buddy that's not even closr to the real answer!!!

21. dan815

maybe some error somewhre xD

22. YumYum247

23. YumYum247

we're looking for the horizontal displacement of the ball....

24. YumYum247

so we figure out the time and plug that into the horizontal equation.... D = t/v or something like that....

25. YumYum247

here is the solution......wait

26. YumYum247

27. YumYum247

actually this topic is slightly different, its about the chaning velocity and the angle....and to find the horzontal displacement we use the formula Dh = Vi2 X sin2O/a but i wanna know what is the Vi means here....

28. YumYum247

is it the initial horizontal velocity or the initial vertical velocity??????

29. dan815

dont use formulas just learn it

30. YumYum247

because i'm pluging in the number and both of my x and y comps are different :( cuz when i ply them into the equation i get a differentt answer :(

31. dan815

i think the answer there is wrong

32. dan815

it should be 123m ish

33. YumYum247

i wanna know what is the Vi2 means here????

34. YumYum247

vertical or horizontal......

35. dan815

what are u referring to again

36. YumYum247

hold up.....le me draw for you :)

37. dan815

wait i found the mistake they made

38. dan815

i think they plugged in wrong numbers or something because they ahve wrong horizontal and veritcal components

39. YumYum247

|dw:1439952832199:dw|

40. dan815

yes that is the right one

41. YumYum247

those are my numbers...when i plug them into their equation, i don't get the right answer????

42. dan815

did u get the method i was showing u up there?

43. YumYum247

i've tried both but i ain't getting 40m :"(

44. dan815

45. YumYum247

yuss i already know that method.....

46. dan815

123 is right

47. YumYum247

you sure bra???

48. dan815

yes

49. dan815

i did it twice so it has to be right

50. YumYum247

well what do you get when you pplug in your findings into the formula, do you get 40?

51. YumYum247

let's try

52. dan815

i dont use formulas i dont know, id have to derive it

53. dan815

what formula do u want me to derive?

54. dan815

u want a formula with velocity and distance only?

55. dan815

:)

56. YumYum247

change in horzontal displacement = Vi^2 X Sin2 Theta/a

57. dan815

okay

58. dan815

|dw:1439953379986:dw|

59. YumYum247

normally i label my vertical values and horizontal values.... Vertical: Horizontal: a = -9.8m/sec2 dh = ? Dh = 0m/s Vh = 26.81m/sec Vi = 22.49m/s up t = ? Vf = 35m/sec t = ?

60. dan815

okay that is fine

61. dan815

that is good

62. dan815

now look maybe u can help work this out

63. dan815

wait why is there a Vf=35 that doesnt eexist

64. dan815

Vertical Vi=22.49m/s up Vf=22.49m/s down

65. dan815

now tell me how much time that will take with 9.8m/s^2 acc

66. YumYum247

i just need to find time, so for that i can use the following dy = Vi X t + 1/2(aXt^2) i'll get a qudratic expression and fro mthere i get my time....i ply that time into the horizontal equation = V = d/t => "t = d/V"

67. dan815

68. dan815

Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf?

69. dan815

remember acceleration =V/t right

70. dan815

so what is the time?

71. YumYum247

a = V/t yes!! :D

72. dan815

so how much time will it take "Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf? "

73. YumYum247

1 sec?

74. dan815

no

75. dan815

okay how much total change was there in the velocity

76. dan815

|dw:1439953983112:dw|

77. YumYum247

wait we have 2 velocities here Vi and Vf so which one am i going to use for that....??

78. dan815

|dw:1439953996826:dw|

79. dan815

accerelation = change in velocity / change in time

80. YumYum247

OMG i got it..... LOL

81. dan815

so how much did the time change by?

82. YumYum247

|dw:1439954092459:dw|

83. dan815

okay good

84. dan815

so this means it took 4.5 seconds for it go up and down

85. YumYum247

d= t/v

86. dan815

in this same time it was constantly moving horizontally at some speed

87. dan815

Horizontal speed * ~4.5 = distance in the horizontal v*t=d

88. YumYum247

pellet...-_-

89. dan815

its oka if the answer does not match the one in the book because that one is wrong, i can show u with a simple thought experiment

90. YumYum247

so 4.5s X 26.81m/s

91. dan815

yeah

92. YumYum247

121 LOL Yus!!! :"D

93. dan815

go online and look at a football being kicked in the air, see the approximate time

94. dan815

it just cannot be 40m meters that is too close

95. dan815

they will fire him if tahts the best he can kick it

96. dan815

this is an opening kick off

97. YumYum247

no but the thing is that, this chapter covers chaning angle and velocity...so to find the hori\ontal displacement .....they gave us the formula for that.....

98. YumYum247

i was just following the steps....

99. dan815

i see i see i get it

100. dan815

i wouldnt worry about all that stuff trust me, you should check out the MIt physics lectures they teach kinematics properly

101. dan815

There is only 2 real formulas u need to know for this whole course

102. dan815

d(t)= displacement as a function of time d'(t)= velocity, first derivative of displacement is velocity d''(t)=acceleration, 2nd der of displament is acc from these you can get all of kinematics equations

103. YumYum247

so what do i do in the exams cuz this chapter and the previous chapter literally had the same types of questions....where you have the final velocity, angle and maybe one or two thingsg from the horizontal.....but virtually exactly the same types of question.....

104. dan815

dont worry, for all those u just need some geometry and these principles

105. dan815

try some more questions, post it on here if u are stuck

106. YumYum247

cuz i picked a question from my previous chapter on "upward movign projectiles" and applied the formula from this chapter" Changing launch angle and initial velocity....the answer i got was totally different,......

107. YumYum247

ok thanks ....:")

108. YumYum247

stay cute!!! :)

109. dan815

wait u got 50 something right

110. YumYum247

yah.....

111. dan815

what u got there is the change in vertical distance

112. YumYum247

nononoono

113. YumYum247

sorry different question.....

114. YumYum247

i'll post it if i have the energy....

115. dan815

okk get to work!

116. YumYum247

eye eye captain....:"D