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  • 一個(gè)物理題.幫幫,

    一個(gè)物理題.幫幫,
    A hot air balloon decends toward the ground with a velocity of (2.0 m/s).A champagne bottle is opened to celebrate takeoff,expelling the cork horizontally with a velocity of (4.0 m/s) relative to the balloon.When opened,the bottle is 6.2 m above the ground.(Neglect air resistance.)
    1.What is the speed of the cork,and its initial direction of motion,as seen by the same observer?
    2.Determine the maximum height above the ground attained by the cork.
    3.How long does the cork remain in the air?
    (2.0 m/s)y (4.0 m/s)x
    英語人氣:680 ℃時(shí)間:2020-05-30 17:01:03
    優(yōu)質(zhì)解答
    先是翻譯,我想你應(yīng)該就會(huì)做了(翻譯錯(cuò)了就不是我的事了,本人英語不大好)
    就是說熱氣球2m/s上升,在上升了6.2m的時(shí)候打開一瓶香檳,軟木塞以4m/s的速度相對(duì)香檳水平飛出,求問:
    1)軟木相對(duì)地面的速度
    2)軟木離地最大高度
    3)軟木離地時(shí)間
    附上題解
    1)
    就是普通的速度合成
    v=4i+2j=2√5m/s,方向與水平面成tanθ=1/2
    2)
    斜拋運(yùn)動(dòng),Vx=4m/s,Vy=2m/s,H0=6.2m
    只考慮豎直方向
    h=Vy²/2g=0.2m
    H=H0+h=6.4m
    3)
    同樣只考慮豎直方向
    上升耗時(shí)t1=Vy/g=0.2s
    下降耗時(shí)t2=√2H/g=0.8√2s
    總耗時(shí)T=t1+t2=0.2+0.8√2 s
    關(guān)于3),如果還要計(jì)算在打開香檳之前的時(shí)間的話,那么要再加上t3=6.2/2 s=3.1s
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