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anonymous

  • one year ago

The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate

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  1. Michele_Laino
    • one year ago
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    your problem is a problem on a parabolic motion. Now your motion can be represented as below: |dw:1439276629065:dw|

  2. Michele_Laino
    • one year ago
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    now, the horizontal component of the velocity is constant, and in general we can write this: \[\Large \begin{gathered} {v_x}\left( t \right) = {V_0}\cos \theta \hfill \\ \hfill \\ {v_y}\left( t \right) = {V_0}\sin \theta - gt \hfill \\ \end{gathered} \] where v_x and v_y are the components of the velocity at a generic time t, and g is the gravity

  3. anonymous
    • one year ago
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    Okayyy will update you ma'am/sir if I have more question!

  4. anonymous
    • one year ago
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    I got confuse

  5. anonymous
    • one year ago
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    Sooooooo I assume y-component will change but x-component will not because the acceleration goes only up to down

  6. rajat97
    • one year ago
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    yes you are right that y-component of velocity will change but x-component of velocity will not change as there is no acceleration or force in the x direction

  7. rajat97
    • one year ago
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    and the speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point

  8. rajat97
    • one year ago
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    so you can use the kinematical equation \[v = ut - (1/2) g t^2\] for the y-component of velocity where u=0 and we know g=10m/s^2 or 9.8m/s^2 (whatever is specified in the question or thats just a matter of our assumption) and t=1.2 from this equation you will get the speed with which the projectile will descend in the y direction after time 1.2s and reffering to the previous post, as i said, that "The speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point" the above equation i.e. \[v = ut - (1/2) g t^2\] is used to find 'velocity' in y-direction and you'll get it as a negative number but we want the positive one as we want the velocity in y-direction when the projectile is ascending which we consider to be positive due to our sign convention now, once you get 'v' i.e. the speed on the y-direction or the y-component of the velocity before 1.2 s, you can find the magnitude of velocity by using the formula for magnitude of resultant of two vectors at right angles to each other (if a abd b are two vectors, perpendicular to each other, the magnitude of their resultant will be sqrt.[(a^2)+(b^2)]) solve this part and then i'll help you with the other parts of the question

  9. anonymous
    • one year ago
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    Okay will read everything. I'll come back to you. Thankyou!

  10. rajat97
    • one year ago
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    wel, thanks for the medal!

  11. anonymous
    • one year ago
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    "and the speed with wich the projectile will descend after time 't' is equal to the speed with wich the projectile was ascending before time 't' in the y-direction, once it reaches the highest point" Is this theory? that If the ground is level, the speed up = speed down?

  12. rajat97
    • one year ago
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    that can be proved and 'If the ground is level, the speed up = speed down' is same as that

  13. anonymous
    • one year ago
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    v=ut−(1/2)gt^2 is the formula you stated right? because I thought the formula is equate to y=vo/u(t)-(1/2)gt^2

  14. anonymous
    • one year ago
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    I got -7.056

  15. rajat97
    • one year ago
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    what is vo/ut are you dividing vo by ut??

  16. anonymous
    • one year ago
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    Hmmm I'll use magnitude formula? soooo sqrt(7.056^2+14^2) = 15.68

  17. rajat97
    • one year ago
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    yes

  18. anonymous
    • one year ago
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    no I'm not dividing vo/t lol it means vo or u

  19. rajat97
    • one year ago
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    yes then you are right

  20. anonymous
    • one year ago
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    Okayyyy but I'm still not convinced that you can use the kinematic equation to find the V

  21. anonymous
    • one year ago
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    OH WAIT

  22. anonymous
    • one year ago
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    you mean to say that v and y-component is the same?

  23. rajat97
    • one year ago
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    wohooo! you got it

  24. rajat97
    • one year ago
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    and you can use the kinematical equation for the y component as the projectile is performing uniformly accelerated motion in the y direction;)

  25. anonymous
    • one year ago
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    okayyyy so moving on then. 7.056 is the y-component right?

  26. anonymous
    • one year ago
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    and 14 is the x-component

  27. anonymous
    • one year ago
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    by which means I'll use arctan to get the angle

  28. rajat97
    • one year ago
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    yes you are right but why do you want the angle??

  29. anonymous
    • one year ago
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    \[\arctan(-7.056/14)\]

  30. anonymous
    • one year ago
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    Oh lol

  31. anonymous
    • one year ago
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    I thought it is needed

  32. rajat97
    • one year ago
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    i am the dumbest person in the world i don't know what i was thinking about but i've posted the wron formula for the y-component of the speed of the projectile it should be [v=u-gt] where v is the speed in the y direction or the y-component of the speed

  33. rajat97
    • one year ago
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    i am very very very very very sorry that i wasted your time on such a silly thing i am very sorry

  34. anonymous
    • one year ago
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    Okayyy let's start over lol soooo

  35. anonymous
    • one year ago
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    I'll use 1.2s = t

  36. rajat97
    • one year ago
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    yes

  37. rajat97
    • one year ago
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    you there?

  38. anonymous
    • one year ago
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    Yessss. I rereading the questionss

  39. rajat97
    • one year ago
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    okay

  40. anonymous
    • one year ago
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    okay so I'll use the formula v = u-gt means u=0 g=9.8 t=1.2

  41. anonymous
    • one year ago
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    5.88

  42. rajat97
    • one year ago
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    okay now use the magnitude formula

  43. anonymous
    • one year ago
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    okay sqrt(5.88^2+14^2)

  44. anonymous
    • one year ago
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    15.1847

  45. rajat97
    • one year ago
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    i think you have done something wrong with the y component of velocity it should be v = 0 - 9.8 x 1.2 = -11.76m/s (we want the positive one so 11.76m/s)

  46. anonymous
    • one year ago
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    omg okay sorry.

  47. anonymous
    • one year ago
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    i did v= 0-1/2(9.8)(1.2) lol

  48. rajat97
    • one year ago
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    no matter man :)

  49. anonymous
    • one year ago
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    So okay moving on

  50. anonymous
    • one year ago
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    so angle will not be taken right?

  51. rajat97
    • one year ago
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    yes angle will not be taken

  52. anonymous
    • one year ago
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    So moving on we got the magnitude

  53. rajat97
    • one year ago
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    yes we get it now

  54. anonymous
    • one year ago
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    I'll illustrate something is this correct

  55. rajat97
    • one year ago
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    okay!

  56. anonymous
    • one year ago
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    18.2838 is the magnitude right?

  57. rajat97
    • one year ago
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    wait a second i'll calculate

  58. anonymous
    • one year ago
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    sqrt(11.76^2+14^2)

  59. rajat97
    • one year ago
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    yes thats the correct figure

  60. anonymous
    • one year ago
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    okay okay ill illutrate something

  61. anonymous
    • one year ago
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    |dw:1439310446738:dw|

  62. rajat97
    • one year ago
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    i'll show you the correct thing

  63. anonymous
    • one year ago
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    okaaaaay thank you

  64. rajat97
    • one year ago
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    well, you cannot show it in a graph like this

  65. anonymous
    • one year ago
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    okay. Hmmm bc I got confused we already have a y-component = 14 right?

  66. anonymous
    • one year ago
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    so we need an x-component to get the magnitude

  67. rajat97
    • one year ago
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    no we have the x-component=14 and y component before 1.2s = 11.76

  68. anonymous
    • one year ago
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    Oh okay wait let me clear this up I thought 14 is the y-component bc it is stated as maximum height

  69. anonymous
    • one year ago
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    oh wait hmm

  70. anonymous
    • one year ago
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    i think im wrong

  71. anonymous
    • one year ago
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    "The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. "

  72. rajat97
    • one year ago
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    look i'll explain you what exactly happens in projectile motion: First of all we need to clear that during the projectile motion, the only force acting on the object is the gravitational force and that too in the vertically downwards direction there is no force in the horizontal direction and we call the vertical direction as the y direction and the horizontal direction as the x direction as there is no force acting on the object in the x direction, the speed of the object will not change in the x direction (Which is 14 here) and the object is performing uniformly accelerated motion in the y direction (you can call it motion under gravity) so we use the kinematical equations for uniformly accelerated motion if we want to find any of the parameters related with the y direction and we use the formula displacement=velocity x time and it's other forms to find any parameters related to the horizontal i.e. the x direction

  73. rajat97
    • one year ago
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    at maximum height, just as motion under gravity, the speed of the object in the y direction is 0 so the speed contributing to the total speed of the object at the heighest point of the projectile motion is the speed in the x direction and here it is 14

  74. anonymous
    • one year ago
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    OKAY GOT IT.

  75. anonymous
    • one year ago
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    Thank you now we alraedy clear it up

  76. rajat97
    • one year ago
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    okay! do you need help with the other parts of the problem??

  77. anonymous
    • one year ago
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    So we got the magnitude

  78. anonymous
    • one year ago
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    I think yes?

  79. anonymous
    • one year ago
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    a.) is the magnitude right the 18.2838

  80. anonymous
    • one year ago
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    same as b.) because of the theory

  81. anonymous
    • one year ago
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    that when the ground is level. they're equal to each other

  82. rajat97
    • one year ago
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    yeah exactly you got it!!!!

  83. anonymous
    • one year ago
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    same as b.) but it will become negative bc it is accelearting downward?

  84. anonymous
    • one year ago
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    or nah

  85. rajat97
    • one year ago
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    it is accelerating downwards in both the cases

  86. anonymous
    • one year ago
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    u sure I thought a.) means upward because it is stated it is before reaching max height, which i thought it is still going up to reach its destination

  87. anonymous
    • one year ago
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    and b.) is down bc it had reach the max height means it will freefall now

  88. rajat97
    • one year ago
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    my internet connection has some problems so i may seem to be offline but i'm not look, the force acting on the object is the force of gravitation which always acts in the downward direction. we know that the acceleration is always directed in the direction of the force thus the acceleration is always downwards if the acceleration would upwards, the projectile would fly up, up and up and never come back but the speed and the acceleration of the object are directed in opposite directions when the object is rising. thus the speed of the object decreases and finally becomes '0' at the MAXIMUM HEIGHT

  89. anonymous
    • one year ago
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    omgg im getting it now

  90. anonymous
    • one year ago
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    I'm so dumb I realize it earlier but I forgot

  91. anonymous
    • one year ago
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    As an object moves upward, the vertical component of its velocity decreases 9.8 m/s each second. At the maximum height, the vertical velocity = 0 m/s

  92. anonymous
    • one year ago
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    but when it moves downward the vertical velocity increases 9.8 m/s each second, until the object hits the ground.

  93. rajat97
    • one year ago
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    yipeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee you got it!!!!

  94. anonymous
    • one year ago
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    god learning is fun hahahaha

  95. anonymous
    • one year ago
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    so what's next is 18.2838 the answer for a and b ? if so. we move to c,d,e,f if you want.

  96. anonymous
    • one year ago
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    Heyyyy it is 1:08AM I'm a sluggish and sleepy we should continue tomorrow for more learning!! Thankyouuuuu

  97. rajat97
    • one year ago
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    yeah and specially learning physics is funnnn!

  98. anonymous
    • one year ago
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    but I think c,d,e,f will be a piece of cake right? Lol

  99. rajat97
    • one year ago
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    no problem !

  100. anonymous
    • one year ago
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    Thank you and God bless!! Sorry if I'm hard to teached. I'm just homeschooling atm with the subject.

  101. rajat97
    • one year ago
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    no you are not hard to teach this happens with everyone because this is a bit confusing! meet you tomorrow

  102. anonymous
    • one year ago
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    1 Attachment
  103. anonymous
    • one year ago
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    This is the illustration right?

  104. anonymous
    • one year ago
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    So moving I'll try to understand C,D,E,F

  105. anonymous
    • one year ago
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    moving on*

  106. anonymous
    • one year ago
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    I'm attempting to finish the problem right now. Good luck to me lol

  107. anonymous
    • one year ago
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    atm im attempting to get the max height

  108. anonymous
    • one year ago
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    Vy(1.4) = 11.76m/s = Vyi - g*1.2 = Vyi - 9.8*1.2 => Vyi = 11.76+11.76 = 23.52m/s up at launch.

  109. anonymous
    • one year ago
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    Vy(1.2)*

  110. anonymous
    • one year ago
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    to get the the t of max height 23.52m/s/9.8m/s^2 = 2.4

  111. anonymous
    • one year ago
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    t=v/a

  112. anonymous
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    ill use the vot-1/2gt = y to get the max height am I on the right track?

  113. anonymous
    • one year ago
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    which is 28.224

  114. anonymous
    • one year ago
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    my final answer for (c,d,e,f) is (-16.8,-7.056,16.8,-7.056)

  115. anonymous
    • one year ago
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    idk if I'm right

  116. rajat97
    • one year ago
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    First of all, thanks for the medal @peachpi well, @arvnoodle , the graph that you have uploaded is correct for the position of the particle but not for it's velocity. Next, you want to use the formula y = vo(t) - (1/2)gt^2 and not "vot-1/2gt = y" and we do not need to find the maximum height (The particle or the object is currently at the maximum height or the topmost point of the projectile) by using the formula y = vo(t) - (1/2)gt^2 , you'll get the position of the particle 1.2 s before it was at the max. height, if you put vo(initial component of velocity in the y-direction which is 0 at the topmost point) = 0, g = 9.8 and t = -1.2(time is negative as we have to find the past situation) you'll get a negative value, which suggests that the particle was below the position to that it is now on and for the x direction, you should use the formula x=vt and put v = 14 and t = -1.2

  117. anonymous
    • one year ago
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    All right then.

  118. anonymous
    • one year ago
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    I got the correct answer @rajat97 woooooo everything is good. Thank you!!

  119. rajat97
    • one year ago
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    oh! it's my pleasure to help you but did you understand whatever i said??

  120. anonymous
    • one year ago
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    Yess!!

  121. rajat97
    • one year ago
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    to solve projectile motion problems, always break the motion into x and y directions and then solve for them individually this may be lenghty but way too better than those projectile motion formulae!

  122. anonymous
    • one year ago
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    I take that as an advice sir. Will do!!

  123. anonymous
    • one year ago
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    I'm off finding more problemsss. Later!

  124. rajat97
    • one year ago
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    please don't call me sir

  125. rajat97
    • one year ago
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    and whenever you need help, please do remember me or there are lots of genius gems that openstudy has got!

  126. rajat97
    • one year ago
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    okay then, bye i gotta go

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