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

1. YumYum247

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2. YumYum247

@IrishBoy123 Police help me!!! :)

3. YumYum247

@dan815

4. YumYum247

@dan815 Dan my love come rescue me from this terrible nightmare HUN!!!! :"D

5. YumYum247

Any idea?????????????????//

6. dan815

not sure, the question is a little too vague for me

7. dan815

does it mean launched by compression of spring or was it just thrown up from equilibriium position

8. dan815

the spring i take it is not attached to the ground

9. YumYum247

don't worry about the compression or anything, just work with the given variables...

10. dan815

im trying to determine what is given

11. YumYum247

the g = -9.81m/sec^2, Vi = 2.3m/sec, angle = 78deg

12. dan815

thats just a projectile problem it has nothing to with spring

13. YumYum247

but we have to solve the problem in projectile sense of view....i think.

14. dan815

oh theres nothing to that then just find V_y

15. dan815

Vy= sin78* 2.3 m/s a_y= -9.8m/s^2 h=? solve

16. dan815

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17. YumYum247

and solve for what????

18. dan815

height, make sure u understand what is going on

19. dan815

it should be natural once u understand what is going on

20. dan815

you have an inital velocity, and ur acceleration is tellng u how much of that velocity u are losing per second

21. dan815

when the object is at its peak height, its veloicty is 0, so we want to see the time it has been travelling under that acceleration for

22. YumYum247

ok but i can't see the picture you just posted properly, some of the content is blocked/missing..... :(

23. dan815

once u get the time its been travelling, you can solve for the actualy distance it travelled in that time

24. dan815

zoom out to see everything

25. dan815
26. YumYum247

ok...bit one more thing, there is a formula for this but i'm sure if that would work...let me post it...hold up :)

27. YumYum247

H=u2sin2(θ)2g

28. YumYum247

it looks something like this, le me draw.

29. YumYum247

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30. YumYum247

is this right???????????

31. dan815

im thinking that is from equating the kinetic and potential energy together

32. dan815

see if both ways gives the same answer if it does then its right

33. YumYum247

but what if it doesn't???:O

34. dan815

1/2 m * V_Y^2 = mgh therefore V*sin ^2 (78)/2g = h

35. dan815

i think you are relying on formulas too much, you have to really understand physics or it wont work

36. YumYum247

ok what to do for now???? which one tho????

37. dan815

check both

38. YumYum247

aight.....!!! Thanks!!! :"D

39. dan815

i know whats gonna happen both ways, but i want you to explore

40. YumYum247

ok dear, i'll try to explore. Thank!!

41. YumYum247

what is the formula for projectile motion, i didn't get it for a sec.... t = Vf +Vi + 1/2(G+t^2)

42. YumYum247

@dan815

43. dan815

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44. dan815

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45. dan815

u can do this integral too to see if gives u the same distance

46. dan815
47. YumYum247

nooooooooooooooooooooooooooooooooooooooi 2 is not the power.....

48. dan815
49. YumYum247

2 multiplies with the sign...i have the formula sheet with me rn.

50. dan815

51. dan815

the formula is wrong

52. dan815

i showed u how the formula is derived

53. dan815

it comes from kinetic1+potenrial1= kinetic2+ potential2

54. YumYum247

ok......

55. dan815

so in the beginning theres only kinetic as the height is 0 and in the end theres only potential as the velocity in y direction is 0

56. dan815

1/2mv^2=mgh m cancel both sides 1/2 V^2=gh so h=1/2v^2/g

57. YumYum247

yah makes sense......

58. dan815

or h=v^2/2g

59. dan815

now v^2 is the V_y^2so

60. YumYum247

wait what do you mean by Vy tho...vertical velocity O_o

61. dan815

V_y^2=(Vi*sin 78)^2 =Vi^2 * sin^2 (78)

62. dan815

yes Vy is the vertical velocity

63. dan815

Vi is more like the speed or the veloicty in that specific direction

64. YumYum247

ok thanks a lot.....i mean alot. :)

65. dan815

okay but u better understand this stuff

66. dan815

the most important thing is you are able to visualize and work through it in your head

67. dan815

or paper

68. YumYum247

yah i'm doing it....it takes time bro, gimme sumtime bro.......

69. YumYum247

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70. YumYum247

@dan815 i'm stuck!!!! :"(

71. dan815

what kinda equation is that

72. dan815

its h(t) = 0 + v_y*t+1/2 * a*t^2

73. dan815

h(t) is notation for h as a function of t

74. dan815

Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2

75. YumYum247

i didn't post the whole thing but i followed your steps.....

76. YumYum247

Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2 if you simplify this you get something i posted.....

77. dan815

\[V_y= vi*sin (78)\\ t=V_y/9.8 \\ h= 0 + v_y*t+1/2 * a*t^2\]

78. YumYum247

79. YumYum247

if you rearrange it.

80. dan815

its not a quadratic u know what t is already u plug that in

81. YumYum247

omg...i didn't see that....sory....:) let me try again....

82. dan815

Vi=2.3 Vy=2.3*sin(78) t=Vy/9.8 h=0+Vy*t + 1/2 a*t^2

83. YumYum247

don't say anything just let me take a look at this with my eyes and brain open...

84. YumYum247

i got it....

85. YumYum247

Thanks!!!

86. dan815

check with h=Vy^2/(2g)

87. YumYum247

yah that works for me....

88. YumYum247

h = vertical velocity/2g gives me the right answer....

89. YumYum247

Anyways don't stress yourself..... go watch a porn to cool off a lil.... :"D i'll get this done on my own now!! :)

90. YumYum247

btw the example you gave me on wolframe didn't have the time input for T.

91. YumYum247

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92. YumYum247

@dan815 please take a look at this.

93. YumYum247

i have to use quadratics formula here....

94. dan815
95. YumYum247

yes i get that...but what happened at the very last step....

96. YumYum247

idk if the website used quadratics formula, or what to get the answe!!!

97. YumYum247

im stuck at the very last part....

98. YumYum247

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99. YumYum247

never mind i got it....

100. YumYum247

Thanks !!