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Mathhelp346

  • 2 years ago

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

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  1. Mathhelp346
    • 2 years ago
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  2. Mathhelp346
    • 2 years ago
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    Activity 1 is the first picture

  3. bhaskarbabu
    • 2 years ago
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    what is the velocity of projection???

  4. Mathhelp346
    • 2 years ago
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    1 Attachment
  5. Mathhelp346
    • 2 years ago
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    Does that help?

  6. bhaskarbabu
    • 2 years ago
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    yep

  7. bhaskarbabu
    • 2 years ago
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    do u know the equations of motion??

  8. Mathhelp346
    • 2 years ago
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    What do you mean?

  9. bhaskarbabu
    • 2 years ago
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    okay leave it

  10. Mathhelp346
    • 2 years ago
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    Ok

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

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

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

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

  15. Mathhelp346
    • 2 years ago
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    No

  16. bhaskarbabu
    • 2 years ago
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    u r of which standard??

  17. Mathhelp346
    • 2 years ago
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    ?

  18. bhaskarbabu
    • 2 years ago
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    class of study?

  19. Mathhelp346
    • 2 years ago
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    8th grade?

  20. bhaskarbabu
    • 2 years ago
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    do u know about displacement,velocity,acceleration etc

  21. Mathhelp346
    • 2 years ago
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    Yes

  22. bhaskarbabu
    • 2 years ago
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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

  23. Mathhelp346
    • 2 years ago
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    Ok

  24. bhaskarbabu
    • 2 years ago
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    now ur first question maximum height of the ball

  25. bhaskarbabu
    • 2 years ago
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    at maximum heicht, final velocity is zero

  26. Mathhelp346
    • 2 years ago
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    Ok

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

  28. bhaskarbabu
    • 2 years ago
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    here acceleration is nothing but acceleration due to gravity

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

  30. bhaskarbabu
    • 2 years ago
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    so finally t =U/g

  31. Mathhelp346
    • 2 years ago
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    What do you mean acceleration is nothing?

  32. bhaskarbabu
    • 2 years ago
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    acceleration is equal to acceleration due to gravity in this case

  33. Mathhelp346
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    Ok

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

  35. Mathhelp346
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    Ok

  36. bhaskarbabu
    • 2 years ago
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    so u can do this on remaining planets

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

  38. bhaskarbabu
    • 2 years ago
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    2.02

  39. bhaskarbabu
    • 2 years ago
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    next maximum height reached by the ball

  40. Mathhelp346
    • 2 years ago
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    .85 seconds

  41. bhaskarbabu
    • 2 years ago
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    for what?

  42. Mathhelp346
    • 2 years ago
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    Venus

  43. bhaskarbabu
    • 2 years ago
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    correct

  44. Mathhelp346
    • 2 years ago
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    Ok

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

  46. Mathhelp346
    • 2 years ago
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    Ok

  47. Mathhelp346
    • 2 years ago
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    Wait how do you find the maximum height?

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

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

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

  51. bhaskarbabu
    • 2 years ago
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    ya

  52. Mathhelp346
    • 2 years ago
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    Ok

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

  54. Mathhelp346
    • 2 years ago
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    Ok

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

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

  57. bhaskarbabu
    • 2 years ago
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    ya right 27.32

  58. Mathhelp346
    • 2 years ago
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    Ok

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

  60. Mathhelp346
    • 2 years ago
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    Ok

  61. bhaskarbabu
    • 2 years ago
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    time required to return to planet's surface

  62. bhaskarbabu
    • 2 years ago
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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

  63. Mathhelp346
    • 2 years ago
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    Ok

  64. Mathhelp346
    • 2 years ago
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    So it's the same answer?

  65. bhaskarbabu
    • 2 years ago
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    no

  66. bhaskarbabu
    • 2 years ago
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    wait this is some what lengthy

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

  68. Mathhelp346
    • 2 years ago
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    Ok

  69. bhaskarbabu
    • 2 years ago
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    but the ball should reach the ground

  70. bhaskarbabu
    • 2 years ago
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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

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

  72. bhaskarbabu
    • 2 years ago
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    now it should reach the ground by traveling 3 feet

  73. Mathhelp346
    • 2 years ago
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    So 27 feet

  74. bhaskarbabu
    • 2 years ago
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    no how do u say that?

  75. bhaskarbabu
    • 2 years ago
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    u der?

  76. Mathhelp346
    • 2 years ago
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    Yes

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

  78. bhaskarbabu
    • 2 years ago
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    t = 0.116 seconds

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

  80. bhaskarbabu
    • 2 years ago
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    hope u understood

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

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

  83. bhaskarbabu
    • 2 years ago
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    it should travel 3 feet downward to reach the ground

  84. Mathhelp346
    • 2 years ago
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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?

  85. bhaskarbabu
    • 2 years ago
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    ya

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

  87. bhaskarbabu
    • 2 years ago
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    no

  88. Mathhelp346
    • 2 years ago
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    but the ho for each of them is 3

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

  90. Mathhelp346
    • 2 years ago
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    So how long do the other ones travel?

  91. bhaskarbabu
    • 2 years ago
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    same 3 feet

  92. Mathhelp346
    • 2 years ago
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    So the other ones travel 3 feet downward?

  93. bhaskarbabu
    • 2 years ago
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    |dw:1355813142620:dw|

  94. Mathhelp346
    • 2 years ago
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    Ok

  95. bhaskarbabu
    • 2 years ago
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    i think now u can understand clearly

  96. bhaskarbabu
    • 2 years ago
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    okay?

  97. Mathhelp346
    • 2 years ago
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    Ok thank you

  98. bhaskarbabu
    • 2 years ago
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    welcome

  99. bhaskarbabu
    • 2 years ago
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    :)

  100. Mathhelp346
    • 2 years ago
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    Mercury's time would be 4 seconds?

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