资源简介 (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 Schoole7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D平抛运动的临界问题PART 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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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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=0.8mv0L=0.6mR=0.2me7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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v2L-R= v1 tL+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=0.4sh=0.8mL=0.6mR=0.2mv1v2e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D平抛运动的临界问题2v1v212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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平抛运动的临界问题2v1v212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?dd=0.4m点击观看视频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平抛运动的临界问题3v0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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平抛运动的临界问题3v1v2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=2.5mh=2md=3ml=9me7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D平抛运动的临界问题3v1v2球网边界线击球点H=2.5mx1y1h=2md=3ml=9m设最大速度为v2,竖直方向:H=gt22d+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 vv1H-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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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 TWO2e7d195523061f1c0006c30382f9a94fcf40f581039171ee7152C150C6C764778087C18C5583B9AEF1266D03E34B2F635197C86C67CE775F9591DBF5647AA59104A0B201FCDA87935B22C0FABB54C018C98031B70871257772CB4502CFD6E1CB41348B3771EFE2400DCB884C85E0469312866D82F00B9E8CF80D69D27AFAC8B47E11013348EFEADDE类平抛运动e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D平抛运动条件:v0a研究方法:与初速度垂直类类平抛运动e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D平抛运动条件:v0a研究方法:与初速度垂直类(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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D1.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e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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 =_____v0v0tatat2总结e7d195523061f1c09e9d68d7cf438b91ef959ecb14fc25d26BBA7F7DBC18E55DFF4014AF651F0BF2569D4B6C1DA7F1A4683A481403BD872FC687266AD13265C1DE7C373772FD8728ABDD69ADD03BFF5BE2862BC891DBB79E950753400A011C7C3B99A25F820103BC51EA757209CB54CD5BE620EDB56A7486908F9B9260006C3940C1CCC47E83D4F0E18A2025A59CCA2D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 展开更多...... 收起↑ 资源预览