Projectile Motion 4 Answers

 

1) Use the following two kinematics equations to show that the above projectile follows a parabolic path.
      

     Δy = vi Δt + 1/2gΔt2

     Δx = vaveΔt

 
     vH = Δx/Δt = vRcosθ
     Δt = Δx/(vRcosθ)
   

     Δy = viΔt + 1/2gΔt2 = vRsinθΔt - 1/2gΔt2

     
     Substituting Δt = Δx/(vRcosθ) yields:
 
     Δy = vRsinθ•Δx/(vRcosθ) - 1/2gΔx2/(vR2cos2θ)
 
     Δy = vRsinθ•Δx/(vRcosθ) - 1/2gΔx2/(vR2cos2θ)
 
     Δy = tanθ•Δx - gΔx2/(2vR2cos2θ)
     
     Δy = ax - bx2
 
     where a = tanθ and b = -g/(2vR2cos2θ) are constants in the absence of air resistance.
 
2) A space-flight projectile is launched in a parabolic trajectory. An astronaut in this practice capsule feels
    weightless.
    (a) Why?
          In the absence of air resistance, a projectile is in free fall because the only force acting on it is its
          weight, Fw.
    (b) In what sense is he weightless?
         He is weightless in the sense that he could not exert a force on a scale because both he and the scale
         are accelerating at the same rate, -g. Although his apparent weight is zero due to his vertical
         acceleration he is not truly weightless. If he was truly weightless, then as a result of his launch, he
         would continue to move in a straight line at constant speed indefinitely in the absence of air resistance
         or any external force.

 

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