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致穎 發問時間: 科學其他:科學 · 3 年前

高中物理題目鉛直上拋?

1.從頂樓鉛直上拋一石,若經頂樓上方h公尺石速率為經過頂樓下方h公尺時的一半,則此物可爬升之最高點具頂樓為幾公尺

2.甲乙兩球分別以V,2V的初速,同時同地鉛直上拋,重力加速度g,當甲球升到最高點瞬間,甲乙兩球之間的高度差為多少

3將兩球同時同地鉛直上拋,若兩球上升最大高度差為35公尺,落地時間差2秒,重力加速度g=10m/s,則兩球拋出的初速度差為多少m/s

1 個解答

評分
  • 3 年前
    最佳解答

    Q.1.

    Take upwards be the positive direction

    Let v m/s be the velocity of the stone at the height h m above the building, then -2v m’s is the velocity of the stone at the height h m below the building.

    Consider the stone from the position of h m above the building to h m below the building

    By using v^2 = u^2+2as

    (-2v)^2 = v^2+2(-g)(-2h)

    4v^2 =v^2+4gh

    3v^2 = 4gh

    v^2 = 4gh/3

    Consider the maximum displacement be s m

    Start from the stone at the height h m above the building to the maximum level

    By using v^2 = u^2+2as

    0^2 = v^2+2(-g)(s – h)

    0 = v^2-2(g)(s – h)

    s – h = v^2/2g

    s = v^2/2g + h

    s = 4gh/3(2g) + h (As v^2 = 4gh/3)

    s = 5h/3 m

    Q.2.

    Take upwards be the positive direction

    Let s1 m be the maximum height of ball A reached, then s2 m is the height of ball B when ball A reaches its maximum height

    For Ball A

    By using v^2 = u^2+2as

    0^2 = v^2+2(-g)(s1)

    0 = v^2-2gs1

    s1 = v^2/2g m

    Time for Ball A reaches its maximum height

    By using s = 0.5(u + v)t

    S1 = 0.5(v + 0)t

    v^2/2g = 0.5vt

    t = v/g sec

    For the height of Ball B travels through this time period

    By using s = ut + 0.5at^2

    S2 = (2v)(v/g) + 0.5(-g)(v/g)^2

    S2 = 3v^2/2g m

    So, the difference in height

    = 3v^2/2g - v^2/2g

    =2v^2/2g

    = v^2/g m

    Q.3

    We can use Q.2. Ball A as the reference

    Maximum height s = v^2/2g m

    And time to reach maximum height t = v/g sec

    So time to return the ground t = 2v/g sec

    Let s1 and s2 be the maximum height of ball 1 and ball 2 respectively

    Let v1 and v2 be the initial velocities of ball 1 and ball 2 respectively,

    While t1 and t2 be the times of ball 1 and ball 2 to return the ground respectively. Given that ball 2 is faster than ball 1.

    S1 = v1^2/2g and S2 = v2^2/2g

    So s2 – s1 = v2^2/2g - v1^2/2g = 35

    v2^2 – v1^2 = 70g = 700 ………….. (1)

    t1 = 2v1/g and t2 = 2v2/g

    so t2 – t1 = 2v2/g - 2v1/g = 2

    v2 – v1 = g = 10 ……………. (2)

    From (1)

    v2^2 – v1^2 = 70g = 700

    (v2 – v1)( v2 + v1) = 700

    v2 + v1 = 70

    v2 + v2 + 10 = 70 (By (2))

    v2 = 40 m/s , so v1 = 30 m/s

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