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Mathhelp346

PLEASEE help with attached: How do I complete the table shown on the second picture?

  • one year ago
  • one year ago

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  1. Mathhelp346
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    • one year ago
  2. Mathhelp346
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    Activity 1 is the first picture

    • one year ago
  3. bhaskarbabu
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    what is the velocity of projection???

    • one year ago
  4. Mathhelp346
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    • one year ago
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  5. Mathhelp346
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    Does that help?

    • one year ago
  6. bhaskarbabu
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    yep

    • one year ago
  7. bhaskarbabu
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    do u know the equations of motion??

    • one year ago
  8. Mathhelp346
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    What do you mean?

    • one year ago
  9. bhaskarbabu
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    okay leave it

    • one year ago
  10. Mathhelp346
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    Ok

    • one year ago
  11. bhaskarbabu
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    time taken to reach the maximum height t = u/g

    • one year ago
  12. bhaskarbabu
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    on earth surface , t = 24/32 =3/4 second

    • one year ago
  13. Mathhelp346
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    So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?

    • one year ago
  14. bhaskarbabu
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    no do u know the formula V = U + gt

    • one year ago
  15. Mathhelp346
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    No

    • one year ago
  16. bhaskarbabu
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    u r of which standard??

    • one year ago
  17. Mathhelp346
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    ?

    • one year ago
  18. bhaskarbabu
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    class of study?

    • one year ago
  19. Mathhelp346
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    8th grade?

    • one year ago
  20. bhaskarbabu
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    do u know about displacement,velocity,acceleration etc

    • one year ago
  21. Mathhelp346
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    Yes

    • one year ago
  22. bhaskarbabu
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    okay so now u learn this formula V= U + at V= final velocity of the body U= initial velocity of the body a = acceleration t = time

    • one year ago
  23. Mathhelp346
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    Ok

    • one year ago
  24. bhaskarbabu
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    now ur first question maximum height of the ball

    • one year ago
  25. bhaskarbabu
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    at maximum heicht, final velocity is zero

    • one year ago
  26. Mathhelp346
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    Ok

    • one year ago
  27. bhaskarbabu
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    so, 0=U+at t = -U/a

    • one year ago
  28. bhaskarbabu
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    here acceleration is nothing but acceleration due to gravity

    • one year ago
  29. bhaskarbabu
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    when a body is projected vertically upwards acceleation due to gravity is taken negetive

    • one year ago
  30. bhaskarbabu
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    so finally t =U/g

    • one year ago
  31. Mathhelp346
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    What do you mean acceleration is nothing?

    • one year ago
  32. bhaskarbabu
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    acceleration is equal to acceleration due to gravity in this case

    • one year ago
  33. Mathhelp346
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    Ok

    • one year ago
  34. bhaskarbabu
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    so finally time required to reach maximum height t = U/g = 24/32 =3/4 second

    • one year ago
  35. Mathhelp346
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    Ok

    • one year ago
  36. bhaskarbabu
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    so u can do this on remaining planets

    • one year ago
  37. Mathhelp346
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    So Mercury would be 2 seconds right

    • one year ago
  38. bhaskarbabu
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    2.02

    • one year ago
  39. bhaskarbabu
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    next maximum height reached by the ball

    • one year ago
  40. Mathhelp346
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    .85 seconds

    • one year ago
  41. bhaskarbabu
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    for what?

    • one year ago
  42. Mathhelp346
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    Venus

    • one year ago
  43. bhaskarbabu
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    correct

    • one year ago
  44. Mathhelp346
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    Ok

    • one year ago
  45. bhaskarbabu
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    do the remaining later lt's go for the next question

    • one year ago
  46. Mathhelp346
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    Ok

    • one year ago
  47. Mathhelp346
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    Wait how do you find the maximum height?

    • one year ago
  48. bhaskarbabu
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    for maximum height reached substitute t = U/g in h(t)

    • one year ago
  49. Mathhelp346
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    You said it was t = -U/a?

    • one year ago
  50. Mathhelp346
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    So I substitute the time required to reach maximum height into the quadratic functions?

    • one year ago
  51. bhaskarbabu
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    ya

    • one year ago
  52. Mathhelp346
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    Ok

    • one year ago
  53. bhaskarbabu
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    u will get h(t) = U^2/2g + ho

    • one year ago
  54. Mathhelp346
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    Ok

    • one year ago
  55. Mathhelp346
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    So the answer to the first one would be 27 ft right

    • one year ago
  56. bhaskarbabu
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    now on earth surface, h = (24)^2/2(32) +3 = 12

    • one year ago
  57. bhaskarbabu
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    ya right 27.32

    • one year ago
  58. Mathhelp346
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    Ok

    • one year ago
  59. bhaskarbabu
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    now u can do the remaining let's move to the next question

    • one year ago
  60. Mathhelp346
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    Ok

    • one year ago
  61. bhaskarbabu
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    time required to return to planet's surface

    • one year ago
  62. bhaskarbabu
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    if the ball is projected from the ground time taken to reach the maximum height = time taken to reach the ground from maximum height

    • one year ago
  63. Mathhelp346
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    Ok

    • one year ago
  64. Mathhelp346
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    So it's the same answer?

    • one year ago
  65. bhaskarbabu
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    no

    • one year ago
  66. bhaskarbabu
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    wait this is some what lengthy

    • one year ago
  67. bhaskarbabu
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    now, time taken by the ball to reach the hand of the person is, U/g + U/g = 2U/g

    • one year ago
  68. Mathhelp346
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    Ok

    • one year ago
  69. bhaskarbabu
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    but the ball should reach the ground

    • one year ago
  70. bhaskarbabu
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    there one more property that, if a ball is projected vertically up with certain velocity ,the ball acquire the same velocity when it reaches the original position

    • one year ago
  71. bhaskarbabu
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    so it reaches the hands of the person with a velocity of 24 feet per second

    • one year ago
  72. bhaskarbabu
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    now it should reach the ground by traveling 3 feet

    • one year ago
  73. Mathhelp346
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    So 27 feet

    • one year ago
  74. bhaskarbabu
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    no how do u say that?

    • one year ago
  75. bhaskarbabu
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    u der?

    • one year ago
  76. Mathhelp346
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    Yes

    • one year ago
  77. bhaskarbabu
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    now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t

    • one year ago
  78. bhaskarbabu
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    t = 0.116 seconds

    • one year ago
  79. bhaskarbabu
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    now add this with 2U/g

    • one year ago
  80. bhaskarbabu
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    hope u understood

    • one year ago
  81. Mathhelp346
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    There's a negative in front of (1/2)g And how did you get 3?

    • one year ago
  82. bhaskarbabu
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    that is given because there the ball is projected vertically upward here the ball is coming down

    • one year ago
  83. bhaskarbabu
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    it should travel 3 feet downward to reach the ground

    • one year ago
  84. Mathhelp346
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    \[h(t)=-\frac{ 1 }{ 2 }g t ^{2}+v _{0}t +h _{0}\] So you got the 3 from Earth's quadratic function right?

    • one year ago
  85. bhaskarbabu
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    ya

    • one year ago
  86. Mathhelp346
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    So does that mean they all travel 3 feet downward?

    • one year ago
  87. bhaskarbabu
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    no

    • one year ago
  88. Mathhelp346
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    but the ho for each of them is 3

    • one year ago
  89. bhaskarbabu
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    the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground

    • one year ago
  90. Mathhelp346
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    So how long do the other ones travel?

    • one year ago
  91. bhaskarbabu
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    same 3 feet

    • one year ago
  92. Mathhelp346
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    So the other ones travel 3 feet downward?

    • one year ago
  93. bhaskarbabu
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    |dw:1355813142620:dw|

    • one year ago
  94. Mathhelp346
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    Ok

    • one year ago
  95. bhaskarbabu
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    i think now u can understand clearly

    • one year ago
  96. bhaskarbabu
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    okay?

    • one year ago
  97. Mathhelp346
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    Ok thank you

    • one year ago
  98. bhaskarbabu
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    welcome

    • one year ago
  99. bhaskarbabu
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    :)

    • one year ago
  100. Mathhelp346
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    Mercury's time would be 4 seconds?

    • one year ago
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