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Projectile motion, radii of trajectory

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  1. Jul 19, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-7-19_12-21-46.png
    upload_2017-7-19_12-22-8.png

    2. Relevant equations


    3. The attempt at a solution
    For part (d),
    The curvature radius of trajectory at its
    1) initial point = horizontal range/2 =( v02 sin (2α))/2g
    2) peak = height of ascent/2 = ( v0 sinα)2 /2g

    Is this correct?

    In this problem, the time of ascent is equal to the time of descent.
    Is there anyway to find it out without calculating the time of ascent and the time of descent ?
     
  2. jcsd
  3. Jul 19, 2017 #2
    upload_2017-7-19_14-47-1.png


    vy (t) = v0 sinα - gt
    vx (t) = v0 cosα

    tan (θ) = vy (t) / vx (t) = tan α - gt/(v0 cosα)

    a = g(- ## \hat y##)
    wτ = g sinθ (- ## \hat θ##),
    wn = g cosθ (- ## \hat r##)

    Is this correct so far?
     
    Last edited: Jul 19, 2017
  4. Jul 19, 2017 #3

    cnh1995

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    You can put (vertical) displacement=0 in the y-displacement equation. It's a quadratic equation.
     
  5. Jul 19, 2017 #4
    This will give me the total time of motion and this, too, I will have to calculate.
    I want to show :
    time of ascent = time of descent without doing calculation, just on the basis of physical interpretation of the problem
    Is this possible?
     
  6. Jul 19, 2017 #5

    cnh1995

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    Yes.
    Your x component of velocity doesn't change. And during ascent, you start with non-zero velocity and end up with zero velocity while during descent, you start with zero velocity and end up with the initial velocity. Displacement is same in both the cases and the motion is under the influence of the same force, i.e. gravity.
    The time a body takes to go up to a certain height is equal to the time it takes to free fall from the same height.
     
  7. Jul 19, 2017 #6

    cnh1995

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    You need to consider the centripetal forces responsible for the curvature here.
    Which force is responsible for the curvature when the body is at the peak? How much is that force?
     
  8. Jul 19, 2017 #7

    When the body is at the initial point,

    mv02 /Ri = mg cosα
    Ri = v02/g cosα

    When the body is at the peak,
    m(v0 cosα)2 /Rp = mg
    Rp = (v0 cosα)2/g

    Is this correct?
     
  9. Jul 19, 2017 #8

    cnh1995

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    Yes.
     
  10. Jul 19, 2017 #9
    What about the next question given in post 2?
     
  11. Jul 19, 2017 #10

    cnh1995

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    Correct. So now you can draw the approximate plot of the tangential and normal acceleration.
    You know their values at t=0, t=T/2 and t=T.
     
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