一个铁球由塔顶静止开始释放,最初3s内位移为X1,最后3s内位移为X2.若X2-X1=30m.则
\u4e00\u7269\u4f53\u4ece\u659c\u9762\u9876\u7aef\u6cbf\u659c\u9762\u7531\u9759\u6b62\u5f00\u59cb\u505a\u5300\u52a0\u901f\u76f4\u7ebf\u8fd0\u52a8,\u6700\u521d3s\u5185\u7684\u4f4d\u79fb\u4e3ax1\uff0c\u6700\u540e3s\u5185\u7684\u4f4d\u79fb\u4e3ax2\uff0c\u82e5x1-x2...\u82e5x2-x1=6m,x1:x2=3:7\u6c42\u659c\u9762\u957f
\u89e3\u5f97:x1=4.5m x2=10.5m 4.5m=0.5*a(3s)2 \u89e3\u5f97a=1m/s2
\u6700\u540e\u4e09\u79d2\u4e2d\u95f4\u65f6\u523b\u7684\u77ac\u65f6\u901f\u5ea6\u7b49\u4e8e\u8fd9\u4e09\u79d2\u7684\u5e73\u5747\u901f\u5ea6\uff0c\u53733.5m/s
\u672b\u901f\u4e3a5m/s \u659c\u9762\u957f12.5m
\u7b54\uff1a\u659c\u9762\u957f\u4e3a2.5m\u3002
\u7965\u89e3\uff1a
\u7531\u521d\u901f\u5ea6\u4e3a0\u7684\u5300\u52a0\u901f\u76f4\u7ebf\u8fd0\u52a8\uff0c\u5728\u76f8\u7b49\u7684\u65f6\u95f4\u5185\uff0c\u5176\u4f4d\u79fb\u4e4b\u6bd4\u4e3a1\uff1a3\uff1a5\uff1a7\uff1a9\uff1a11\u3002\u3002\u3002
\u6839\u636e\u9898\u610f\uff0c\u6700\u521d3\u79d2\u4f4d\u79fb\u6bd4\u6700\u672b3\u79d2\u4f4d\u79fb\u4e4b\u6bd4\u4e3a3\uff1a7\uff0c\u5219\uff081\uff0b3\uff0b5\uff09\uff1a\uff085\uff0b7\uff0b9\uff09\uff1d3\uff1a5\u3002
\u518d\u7531\u540e3\u79d2\u4f4d\u79fb\u6bd4\u524d3\u79d2\u4f4d\u79fb\u59271.2m\uff0c\u5373\uff085\uff0b7\uff0b9\uff09x\uff0d\uff081\uff0b3\uff0b5\uff09x\uff1d1.2m,\u6bcf\u4e00\u4efd\u4e3ax=0.1m,\u659c\u9762\u957f\u51711\uff0b3\uff0b5\uff0b7\uff0b9\u4efd\uff0c\u8ba125\u4efd\uff0c\u6240\u4ee5\u4e3a2.5m\u3002
时间中点瞬时速度=平均速度
t1=1.5s 的瞬时速度V1=最初3s内平均速度=X1/3
t2=(t-1.5)s 的瞬时速度V2=最后3s内平均速度=X2/3
而 V2-V1=g(t2-t1)
故 (X2-X1)/3=10(t-3)
即 30/3=10t-30
t=4s
h=(1/2)gt^2=0.5*10*4^2=80m
求加速度是吧，设为a
X1=(at1^2)/2
X2=V1t2+(at2^2)/2
V1=at1
X2-X1=30
得a=0.3
绛旓細鏁 (X2-X1)/3=10(t-3)鍗 30/3=10t-30 t=4s h=(1/2)gt^2=0.5*10*4^2=80m
绛旓細寮濮3绉掞細1/2g3²=S1 鏁翠釜杩囩▼锛1/2gt²=h 鏈鍚3绉掞細h-1/2g(t-3)²=S2 浣嶇Щ鍏崇郴锛歋2-S1=6 浠ｅ叆锛歡=10 瑙ｅ緱锛歵=3.2 S1=45 S2=51 h=51.2s
绛旓細濉旈《绂荤涓灞傜殑楂樺害h2==54.45m 鎵浠ュ鐨勬婚珮搴=h1+h2=61.25m 锛2锛夊皬鐞冧粠濉旈《钀藉埌鍦伴潰鐨勬绘椂闂磘==3.5s 骞冲潎閫熷害=17.5m/s
绛旓細涓鐗╀綋浠庢枩闈㈤《绔部鏂滈潰鐢遍潤姝㈠紑濮嬪仛 鍖鍔犻熺洿绾胯繍鍔,鏈鍒3s鍐呯殑浣嶇Щ涓簒1锛屾渶鍚3s鍐呯殑浣嶇Щ涓簒2锛岃嫢x1-x2... 96 涓涓搧鐞冪敱濉旈《闈欐寮濮嬮噴鏀,鏈鍒3s鍐 浣嶇Щ涓篨1,鏈鍚3s鍐呬綅绉讳负X2.鑻2- X1=30m.鍒 3 涓鐗╀綋浠庢枩闈㈤《绔敱闈欐寮濮嬪寑鍔犻熶笅 婊戝埌鏂滈潰搴曠,鏈鍒3s鍐呯殑浣嶇Щ涓簒1,鏈鍚 3s鍐呯殑...
绛旓細锛1锛夎В锛歨=1/2gt^2=1/2脳10脳4=20m 锛2锛夎В锛歷=gt t=v/g=30/10=3s h=1/2gt^2=1/2脳10脳9=45m 浣犲ソ锛屽緢楂樺叴涓轰綘瑙ｇ瓟锛屽笇鏈涘浣犳湁鎵甯姪锛岃嫢婊℃剰璇峰強鏃堕噰绾炽
绛旓細S=1/2gt鏂 浠ｅ叆80=1/2*10t鏂 寰楀埌t=4绉
绛旓細1/2mv²-1/2mv0²=mgh,v0=15m/s,v鑷繁绠椼傚鏋滄病閫夎繃鐨勮瘽灏辩敤鐗涢】瀹氱悊锛宧=1/2gt²,姹傚嚭钀藉湴鐨勬椂闂磘.鐒跺悗v绔栫洿=gt锛寁=鏍瑰彿涓媣绔栫洿²+v0²锛2锛墄=v0t锛宼鍦ㄧ涓闂凡缁忔眰鍑烘潵浜 锛3锛墆=80m锛屾讳綅绉籹=鏍瑰彿涓媥²+y²锛屾湁涓嶆噦鐨勭户缁棶銆
绛旓細2.BC 3.A 4.A 5.C 6.AC 7.C 8.A 9.B
绛旓細闂鏄粈涔堬紵锛燂紵
绛旓細閲嶅姏鍋氬姛锛800J 闃诲姏鍋氬姛锛-160J 鍏嬫湇闃诲姏鍋氬姛锛160J 鍚堝鍔涘仛鍔燂細640J
