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March 26, 2017

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From a helicopter rising vertically with a velocity of 9 m/s a weight is dropped and reaches the ground in 18 seconds. how high above the ground was the helicopter when the the weight dropped striked the ground.

  • Physics - ,

    Upward motion of the weight
    h =v₀t-gt²/2
    v= v₀ -gt
    v=0 => v₀ =gt => t = v₀ /g =9/9.8 =0.92 s
    h = v₀²/2g =9²/2•9.8 =4.13 m
    Let h₀ is the height of helicopter when the weight began its motion. Then
    for downward motion of the weight
    h₀+h =gt₀²/2
    t₀ =sqrt{2(h₀+h)/g}
    ====
    t₁=2t+ t₀ = 2v₀/g + sqrt{2(h₀+h)/g} =18 s
    sqrt{2(h₀+h)/g} =18 - 2v₀/g
    Square this equation, substitute the given data and calculated height h, and solve for h₀
    h₀ = 1280 m
    During the motion of the weight, the helicopter moved upwards by the distance
    h₁=v₀t₁ = 9•18= 162 m
    Therefore,
    H = h₀ + h₁ =1280+ 162 = 1442 m .

  • Physics - ,

    h1 = Vo*t + 0.5g*t^2
    h1 = -9*18 + 4.9*18^2 = 1426 m Above
    ground when bag was released.

    h2 = 9*18 = 162 m.. Above point of release.

    h = 1426 + 162 = 1588 m. Above ground.

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