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How long is discus in the air and the horizontal distance traveled

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A discus is released at an angle of 42.0° with respect to the horizontal and a velocity of 28.0m/s.

    a)How long does it stay in the air?

    b)What horizontal distance does it travel?

    2. Relevant equations
    dv=1/2*at2
    dh=Vh*Δt
    Kinetic equation d=Vi*t+ 1/2*at2
    dh=-V2*sin2θ /g
    sinθ=opp/hyp
    cosθ=adj/hyp
    Δt=-2Vsinθ / g

    3. The attempt at a solution

    hyp=28.0m/s

    cos(42°)*28.0m/s=Vh= 20.8m/s
    sin(42°)*28.0m/s=Vv= 18.7m/s

    a) I used Δt=-2Vsinθ / g and solved for Δt=3.82s

    b) Used dh=-V2*sin2θ /g to solve for dh= 79.6m

    For part b) I don't know when you I am supposed to use dh=-V2*sin2θ /g equation and when I should just use dh=Vh*Δt or the d=Vi*t+ 1/2*at2 equation. If anyone could give me any tricks or explanations of when and why we use that specific equation that would be greatly appreciated. Thank you so much in advance!
     
  2. jcsd
  3. Apr 14, 2012 #2

    tiny-tim

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    hi dani123! :smile:
    (what's the dh=-V2*sin2θ /g equation? :confused:)

    if you can use dh=Vh*Δt or d=Vi*t+ 1/2*at2 then you certainly should :smile:
     
  4. Apr 15, 2012 #3
    dh=-V2*sin2θ /g equation was given in my course book for finding the horizontal range of a projectile, launched at an angle.
    Where V is the instantaneous projectile velocity,
    θ is the angle of projection, and
    g is the acceleration due to gravity which is equal to -9.80 m/s2
    This equation can be used to find the horizontal range of projectile that returns to the same level from which it was launched (dv=0), provided that angle of projection, θ, and the instantaneous projectile velocity, V, are known.
     
  5. Apr 15, 2012 #4

    tiny-tim

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    ahh! :smile:

    in that case, the answer to your original question …
    … is that both methods work, so you should use the quick method if you can remember it

    (which i'm pretty sure you won't be able to in the exam :wink:)
     
  6. Apr 15, 2012 #5
    Thank you!
     
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