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匿名使用者 發問時間: 科學其他:科學 · 6 年前

F.4 Physics Mechanics

Please help me with this question, and show steps. Thank you!

Two identical balls X and Y are released from rest from the top of a building 100 m above the ground. Ball Y is released 2 s after ball X. What is the maximum vertical separation between balls X and Y in the air? (Neglect air resistance.)

2 個解答

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  • 最佳解答

    Vx stands for ball X 's velocity

    Vy stands for ball Y

    Vxy stands for "velocity of Y relative to X"

    Vxy=Vy-Vx

    Acceleration=g=10 m/s^2

    when the Y is released

    X has displacement=1/2 * 10 * (2^2)=20

    Vxy=Vy-Vx=Vy-0=Vy=10*2=20

    Axy=Ax-Ay=g-g=0 (acceleration of xy)

    as a result Vxy is const =20

    so maximum vertical separation between balls X and Y in the air is

    "when the X touches ground"

    it take

    t=(2*100/g)^1/2

    y needs to fly for t=(2*100/g)^1/2 "-2" (released 2 s after ball X)

    so the answer is 20+(2√5-2)*20=40√5-20

  • LI
    Lv 7
    6 年前

    從位移看:

    Sx = 1/2*gt^2

    Sy = 1/2*g(t-2)^2

    兩者相距

    D = Sx – Sy = 1/2*gt^2 - 1/2*g(t-2)^2 = 2g(t-1)

    所以t越大D越大

    直到X撞地

    100 = 1/2*gt^2

    t = √(200/g)

    所以

    D = 2g(√(200/g)-1) = 2( 10√(2g) – g )

    若g=10

    D= 40√5-20

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