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Projectile motion after a flying helical path

  1. Dec 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A bird flew along a helical path [tex]r(t)=<5cost,5sint,2t>[/tex], at time t=10 second, the bird died instantly. where did it hit the xy-plane ? g=32 ft/s/s.

    2. Relevant equations
    [tex]r(t)=(v_0 \cos\theta t) i+(v_0 \sin\theta t - 0.5gt^2 + h_0) j[/tex]

    3. The attempt at a solution
    taking derivative of r(t), I got the velocity function respect to t, [tex]v(t)=<-5sint,5cost,2>[/tex]
    at t=10 s, we are located at (-4.2, -2.7, 20)
    so that means the velocity in the x-direction at time =10 seconds is 2.72 ft/s
    and the velocity in the y-direction at t=10 s = -4.2 ft/s
    to hit the ground means that the distance in Z direction is 0, setting [tex]0=z(10)+v_z(10) t - 16t^2[/tex]
    velocity in z direction is constant= 2
    we can find the time the bird hit the ground, which = 1.18 second.
    now, plug t and velocity in the x-direction into [tex]X_f=x(10)+v_x(10) t [/tex], we can find x-axis distance=-9.9 ft
    [tex]Y_f=y(10)+v_y(10) t [/tex] similarly, we can find y - coordinate= -76.56 ft ( this value seems way off)

    i am confused, does angle of projection matters in this case? if does, how do we find the angle of projection?
     
    Last edited: Dec 15, 2015
  2. jcsd
  3. Dec 15, 2015 #2

    BvU

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    ##
    X_f=x(10)+v_x(10) t
    = -4.2 + 2.7 * 1.18 ## can never be -9.9 ft !
    For y you simply have a decimal point error, I'd guess: ##\ -2.7 - 1.18 * 4.2 = -7.7##

    Always estimate the answer when using a calculator or a spreadsheet.

    And I wouldn't round off before I had the answer.
     
  4. Dec 15, 2015 #3
    so my approach was correct? thank you
     
  5. Dec 16, 2015 #4

    BvU

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    Approach was correct, execution a bit sloppy. And: You're welcome.
     
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