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Projectile Motion (potential & kinetic energy)

  1. Nov 11, 2012 #1
    The question is in the attachment.

    a) for the height i got 1.25m which you get by using displacement and acceleration. (t=20/40=0.5sec)

    b) a=v/t, 10*0.5=5m/s

    c) At initial point:

    mgh= 0.040*10*1.25=0.5J
    1/2mv^2=1/2*0.040*40^2=32J

    Hits Ground:
    mgh=0.040*10*0=0J
    1/2mv^2=1/2*0.040*(20 or 40)^2 ?????? The answer is 32.5J im not sure what happens here and how to get that. Do you add the energys from initial point together???
     

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    Last edited: Nov 11, 2012
  2. jcsd
  3. Nov 11, 2012 #2

    lewando

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    Since there are no losses due to wind resistance the initial energy equals the final energy. The velocity you use right before it hits ground is the composite velocity.
     
  4. Nov 11, 2012 #3
    sorry i dont understand that very well.
     
  5. Nov 11, 2012 #4

    tiny-tim

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    Hi Arty7! :smile:
    Use components. :wink:
     
  6. Nov 11, 2012 #5
    so.. mgh+1/2mv^2 0.5 +32 =32.5J
     
  7. Nov 11, 2012 #6
    As for b use vy=g*t
     
  8. Nov 11, 2012 #7
    a) Use the formula : h=(1/2)gt²
    h=1[itex]/[/itex]2 [itex]\times[/itex] 10[itex]\times[/itex](0.5)²
    Here I calculated t from the usual t = d[itex]\div[/itex]speed
    t=0.5 s
    Coming back to h,We get h as 1.25 m

    b) v[itex]_{x}[/itex] = u = 40 m/s
    v[itex]_{y}[/itex] = u[itex]_{y}[/itex]+gt=0+10[itex]\times[/itex]0.5
    = 5 m/s

    c) KE[itex]_{initial}[/itex] = (1[itex]\div[/itex]2)[itex]\times[/itex]m[itex]\times[/itex]v²
    m=40 g and v= 40 m/s
    = 0.5 J
    KE[itex]_{final}[/itex] = (1[itex]\div[/itex]2)[itex]\times[/itex]m[itex]\times[/itex]v[itex]_{final}[/itex]²
    Here you need to find the final velocity v[itex]_{final}[/itex] ; Find by v = u + gt
    u=40 m/s and t=0.5 s
    So v = 45 m/s
    ∴ KE[itex]_{final}[/itex] = 0.9 J if I'm right
    Now KE[itex]_{final}[/itex] - KE[itex]_{initial}[/itex] = 0.4 J

    Similarly Potential Energy [itex]\rightarrow[/itex]
    PE[itex]_{initial}[/itex] = mgh
    [itex]\rightarrow[/itex]40 × 10[itex]^{-3}[/itex] × 10 × 1.25
    = 0.5 J
    PE[itex]_{final}[/itex] = mgh
    Here h is the equation of the trajectory,
    We get h as 6.25 × 10[itex]^{-3}[/itex]
    Now the potential energy is 2.5 × 10[itex]^{-4}[/itex] J
    I guess PE(final) - PE(inital) = 2 × 10[itex]^{-4}[/itex] J
    Now PE + KE = Answer
    PE + KE = ?
     
    Last edited: Nov 11, 2012
  9. Nov 11, 2012 #8

    lewando

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    I think you meant:
    v=[itex]\sqrt{v^{2}_{x} + v^{2}_{y}}[/itex]
     
    Last edited: Nov 11, 2012
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