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匿名 發問於 科學及數學其他 - 科學 · 1 十年前

physic 拋物向上求building height

係天台拋一樣野向上 with speed 16ms-1

個樣野會reach the ground after 4seconds

find the height of the building

比個step 我,唔該~~~

更新:

the Answer is 16 m !!!!!

更新 2:

agnissammy and seiya

錯了

3 個解答

評分
  • 1 十年前
    最愛解答

    Divide the journey into two main parts. Take downwards as positive.

    In the first part, the object reaches its maximum height and then returns to original position, which is at the top of the building.

    v2 – u2 = 2as

    ( 16 )2 – ( 0 )2 = 2 ( 10 )( s )

    s = 12.8 m

    s = ut + at2 / 2

    12.8 = ( 0 )( t ) + ( 10 )( t2 ) / 2

    t = 1.6s

    But it is only half of the first part of the journey; hence the time for the 1st part is 3.2s.

    Then the time for the second part of the journey will be 4 – 3.2 = 0.8s and at the start of the second part, the object has an initial v of 16m/s.

    s = ut + at2/ 2

    s = ( 16 )( 0.8 ) + ( 10 )( 0.8 )2 / 2

    = 16 m

    So the height of the building is 16m.

    資料來源: My Phy Knowledge
  • 志仁
    Lv 6
    1 十年前

    s=ut+1/2 a t^2

    a=-g=-9.8ms^-1

    t=4, u=16m/s

    s=16(4)-4.9(16)

    =48-78.4

    =-30.4m

    The building is 30.4m in height.

    The minus sign mean the ground is under your feet at 30.4m.

    2007-07-25 12:52:12 補充:

    a=-g=-9.8ms^-2

  • 1 十年前

    s = ut+ 1/2 (at^2)

    2007-07-25 12:50:22 補充:

    s = height of the buildingu= initial speeda= gravitational accelerationt=time taken

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