1.4专题四平抛运动规律的应用(二)课件+-2022-2023学年高一下学期物理教科版(2019)必修第二册(共27张PPT)

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1.4专题四平抛运动规律的应用(二)课件+-2022-2023学年高一下学期物理教科版(2019)必修第二册(共27张PPT)

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(共27张PPT)
厚德求正 博学求真
1.专题四-平抛运动规律的应用(二)
“从物理学出发思考一切” —— 爱因斯坦
Honghu Senior High School
平抛运动的临界问题
01
类平抛运动
02
目 录
CONTENTS
厚德求正 丨 博学求真
Honghu Senior High School
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平抛运动的临界问题
PART ONE
1
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h
v1
v2
小球从某一高度抛出,速度大小在什么取值范围才能扔进垃圾桶?
平抛运动的临界问题1
点击观看视频
画一画:小球分别落在哪两个位置对应能进桶的最小速度与最大速度?
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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
平抛运动的临界问题1
已知h=0.8m,抛出点距离垃圾桶正中心的水平距离L=0.6m,垃圾桶半径R=0.2m,不计空气阻力,g取10 m/s2 ,求:
(1)小球落入垃圾桶所用的时间t;
(2)小球要落入垃圾桶里,初速度大小的取值范围.
h=0.8m
v0
L=0.6m
R=0.2m
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平抛运动的临界问题1
(2)设小球能落入垃圾桶的最小速度为v1,最大速度为v2
L-R= v1 t
L+R=v2 t
解得 v1 =(L-R)
v2 =(L+R)
代入数据得v1 =1m/s v2 =2m/s
(1)竖直方向上:
h=gt2
解得 t=
代入数据得t=0.4s
h=0.8m
L=0.6m
R=0.2m
v1
v2
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平抛运动的临界问题2
v1
v2
1
2
要打在第2节台阶上,抛出时的速度大小在什么取值范围?
点击观看视频
画一画:小球分别落在哪两个位置对应能打在第2节台阶的最小速度与最大速度?
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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
平抛运动的临界问题2
v1
v2
1
2
已知每节台阶的高度和宽度都是d=0.4m,不计空气阻力,g取10 m/s2 ,要想小球打在第2节台阶上,初速度大小的取值范围?
d
d=0.4m
点击观看视频
e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
平抛运动的临界问题3
v0
球网
边界线
击球点
要想过网,且不出界,击球速度大小在什么取值范围?
画一画:小球分别落在哪两个位置对应能过网且不出界的最小速度与最大速度?
点击观看视频
e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D
平抛运动的临界问题3
v1
v2
球网
边界线
击球点
击球高度H=2.5m,球网高h=2m,距离球网d=3m,球网距离边界线l=9m,不计空气阻力,g取10 m/s2 ,要想既不触网又不出界,球的初速度大小取值范围
H=2.5m
h=2m
d=3m
l=9m
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平抛运动的临界问题3
v1
v2
球网
边界线
击球点
H=2.5m
x1
y1
h=2m
d=3m
l=9m
设最大速度为v2,
竖直方向:
H=gt22
d+l= v2t2
水平方向:
解得 v2 =(d+l)
代入数据得 v2 =12m/s
设最小速度为v1,
竖直方向:
解得 v1 = d
代入数据得 v1=m/s
因此速度的取值范围为m/s v
v1
H-h=gt12
水平方向:
d= v1t1
受下落高度限制,选高度限制下的研究过程
一.平抛运动的临界问题
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笔记时间
Notes !!!
1.临界情况常见字眼:
(1) “刚好”“恰好”“正好”
(2) “取值范围”“多长时间”“多大距离”
(3) “最大”“最小”“至多”“至少”
2.确定临界轨迹:
(1) 受水平位移限制时:
(2) 受下落高度限制时:
类平抛运动
PART TWO
2
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类平抛运动
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平抛运动条件:
v0
(1)水平初速度;
(2)只受重力(a=g,方向竖直向下)
g
研究方法:
运动的分解
水平方向:匀速直线
竖直方向:自由落体运动
竖直向下
水平
与初速度垂直
类平抛运动
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平抛运动条件:
v0
a
研究方法:
与初速度垂直

类平抛运动
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平抛运动条件:
v0
a
研究方法:
与初速度垂直

(1)初速度;
(2)合外力为与初速度垂直的恒力
运动的分解
沿初速度方向:匀速直线
沿合外力方向:初速度为0的匀变速直线
类平抛运动
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有一光滑固定斜面,一物块从斜面左上方顶点P沿水平方向射入,恰好从底端右侧Q点离开,请思考:
物块做什么运动?
直线or曲线?加速度大小、方向?
用什么研究方法来分析这种运动?
点击观看视频
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1.如图所示的光滑固定斜面长为l、宽为b、倾角为θ,一物块(可看成质点)从斜面左上方顶点P沿水平方向射入,恰好从底端右侧Q点离开斜面,已知重力加速度为g,不计空气阻力,求:
(1)物块加速度的大小a;
(2)可以把物块的运动怎样分解;
(3)物块由P运动到Q所用的时间t;
(4)物块由P点水平射入时初速度的大小v0.
典例练习
典例练习-解析
(1)对物块进行受力分析,根据牛顿第二定律:
mgsinθ=ma
解得a=gsinθ
(4)沿水平方向
b=v0t
解得t=
(2)物块沿初速度方向不受力,做匀速直线运动;沿合力方向初速度为0,做匀加速直线运动
(3)沿斜面方向
l=at2
解得v0= b
二. 类平抛运动
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笔记时间
Notes !!!
1. 受力特点:
物体所受合外力为________,且与初速度方向________.
2. 研究方法:
恒力
垂直
将运动分解为沿初速度方向的________________和沿合外力方向的初速度为0的________________.
匀变速直线运动
匀速直线运动
3. 运动规律:
初速度方向上:
合外力方向上:
vx=____
x=_____
a=_____
y=________
vy =_____
v0
v0t
at
at2
总结
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速度反向延长线平分水平位移
tan θ=2tan α
恒力
分解
初速度
垂直初速度

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