一个物理题.帮帮,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 bot
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/26 15:36:22
一个物理题.帮帮,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 bot
一个物理题.帮帮,
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
一个物理题.帮帮,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 bot
先是翻译,我想你应该就会做了（翻译错了就不是我的事了,本人英语不大好）
就是说热气球2m/s上升,在上升了6.2m的时候打开一瓶香槟,软木塞以4m/s的速度相对香槟水平飞出,求问：
1）软木相对地面的速度
2）软木离地最大高度
3）软木离地时间
附上题解
1）
就是普通的速度合成
v=4i+2j=2√5m/s,方向与水平面成tanθ=1/2
2）
斜抛运动,Vx=4m/s,Vy=2m/s,H0=6.2m
只考虑竖直方向
h=Vy²/2g=0.2m
H=H0+h=6.4m
3）
同样只考虑竖直方向
上升耗时t1=Vy/g=0.2s
下降耗时t2=√2H/g=0.8√2s
总耗时T=t1+t2=0.2+0.8√2 s
关于3）,如果还要计算在打开香槟之前的时间的话,那么要再加上t3=6.2/2 s=3.1s