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Find the time of movement

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle is thrown in the angle pi/3 radian with speed 10m/S. Find the time of movement if the maximal height is equal to the distance of fall.

    2. Relevant equations
    v^2-voy^2=2gh

    3. The attempt at a solution
    0-10^2*3/4=2*10*hmax
    hmax=3.75
    v0x*2t=3.75
    v0*cos60*2t=3.75
    10*0.5*2*t=3.75
    t=0.375. The answer in my book is 1S.
    I also want to know why whe I use the formula of maximal height v0^2sin^a/2g and maximal fallL=v0^2sin2a/g with the data from the problem they don't come out to be equal.
     
  2. jcsd
  3. Nov 6, 2016 #2
    Can you explain what "time of movement" is? Is this the total time while the particle is in the air? What is the exact wording of the problem?

    I am getting the same maximum height and using the equation $$ h = \frac{1}{2}at^2 $$ I am getting a total flight time of ## \sqrt{3} ##
     
    Last edited: Nov 6, 2016
  4. Nov 6, 2016 #3

    haruspex

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    There seems to be too much information. That can be fixed by dropping the assumption that it lands at the same height it was launched at, but I still don't get the book answers.
     
  5. Nov 6, 2016 #4

    PhanthomJay

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    The angle of pi/3 radians might be measured from the vertical axis.
     
  6. Nov 6, 2016 #5

    haruspex

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    By dropping the assumption that it lands at the same height it was launched at, using pi/3 as angle to the horizontal I get a time of 1/4 s. Taking it as angle to the vertical makes it even less. To get a time of 1s I need a launch angle of acos(√2-1) to the horizontal, or about 65 degrees. That's using g=10m/s2.
     
  7. Nov 6, 2016 #6

    PhanthomJay

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    Poorly worded problem I can only assume it it is thrown at pi/3 radians with vertical on level ground. Vert comp of velocity thus 5 m/s. Rises to a height h then falls back vert distance h to ground, which is a 2nd assumption. Half sec up and half sec down. Problem should be tossed.
     
  8. Nov 6, 2016 #7

    haruspex

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    Right, so it was just a confusing way of saying it lands back at the height it was thrown from. Well deduced.
     
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