Question

19£®ÈçͼËùʾ£¬¾²Ö¹ÔÚˮƽÃæÉÏÖÊÁ¿Îª2mµÄƽ°å³µAµÄ×ó¶Ë¹Ì¶¨Çᵯ»É£¬´Ëʱµ¯»Éδ·¢ÉúÐα䣬PΪ×ÔÓɶ˵㣬A³µÉϱíÃæ¾àÀëµØÃæ¸ßH=$\frac{{v}^{2}}{6g}$£¬ÖÊÁ¿¾ùΪmµÄ»¬¿éB¡¢C¾²ÖÃÓÚAÉϱíÃ棬ÏÖ½«µ¯»ÉѹËõºó·ÅÊÖ£¬BÍÑÀ뵯»Éºó¾­¹ýʱ¼ätÒ»¶¨»áÓëCÔÚA×îÓÒ¶ËÅöײ£¬Åöײ˲¼ä±»Õ³×¡£¬²¢×îÖÕÂäµ½µØÃæDµã£¬ËÙ¶Èƫת½Ç¦È=30¡ã£¬ÖØÁ¦¼ÓËÙ¶ÈΪg£¬vΪÒÑÖªÁ¿£¬¸÷Á¿²ÉÓùú¼Êµ¥Î»ÖÆ£¬B£¬C¿É¿´×öÖʵ㣬²»¼ÆÒ»ÇÐĦ²Á£®
£¨1£©BCײºóÕûÌåµÄËÙ¶Èv₁£»
£¨2£©µ¯»ÉѹËõºó¾ßÓеÄ×î´óµ¯ÐÔÊÆÄÜE₁£»
£¨3£©BCÂäµØʱÓëPµãµÄˮƽ¾àÀëx£®

Answer 1

·ÖÎö £¨1£©BCÕûÌå×öƽÅ×Ô˶¯£¬¸ù¾ÝƽÅ×Ô˶¯»ù±¾¹«Ê½½áºÏ¼¸ºÎ¹ØϵÇó³öBCײºóÕûÌåµÄËٶȣ»
£¨2£©BCÅöײµÄ¹ý³ÌÖУ¬¶¯Á¿Êغ㣬¸ù¾Ý¶¯Á¿Êغ㶨ÂÉÇó³öÅöײǰBµÄËٶȣ¬ABµ¯¿ª¹ý³Ì¶¯Á¿Êغ㣬¸ù¾Ý¶¯Á¿Êغ㶨Âɼ°ÄÜÁ¿Êغ㶨ÂÉÁÐʽ¼´¿ÉÇó½â£»
£¨3£©·Ö±ðÇó³öBCƽÅ×ˮƽλÒÆ¡¢BÍÑÀ뵯»Éºóʱ¼ätÄÚ·¢ÉúµÄλÒÆÒÔ¼°BÍÑÀ뵯»Éºóµ½BCÂäµØ£¬A¹²·¢ÉúµÄλÒÆ£¬ÈýÕßλÒÆÖ®ºÍ¼´ÎªBCÂäµØʱÓëPµãµÄˮƽ¾àÀ룮

½â´ð ½â£º£¨1£©BCÕûÌå×öƽÅ×Ô˶¯£¬ÔòÓУº
H=$\frac{1}{2}g{{t}_{1}}^{2}$£¬
½âµÃ£º${t}_{1}=\frac{\sqrt{3}v}{3g}$
¸ù¾Ý$tan¦È=\frac{{v}_{y}}{{v}_{0}}=\frac{g{t}_{1}}{{v}_{1}}$µÃ£º
v₁=v
£¨2£©BCÅöײµÄ¹ý³ÌÖУ¬¶¯Á¿Êغ㣬ÉèÅöײǰBµÄËÙ¶ÈΪv₂£¬ÔòÓУºmv₂=2mv₁£¬½âµÃ£ºv₂=2v£¬
ABµ¯¿ª¹ý³Ì¶¯Á¿Êغ㣬É赯¿ªºóAµÄËÙ¶ÈΪv₃£¬¸ù¾Ý¶¯Á¿Êغ㶨Âɼ°ÄÜÁ¿Êغ㶨Âɵãº
2mv₃=mv₂£¬${E}_{P}=\frac{1}{2}¡Á2m{{v}_{3}}^{2}+\frac{1}{2}m{{v}_{2}}^{2}$
½âµÃ£ºv₃=v£¬${E}_{P}=3m{v}^{2}$£¬
£¨3£©BCƽÅ×ˮƽλÒÆ${x}_{1}={v}_{1}{t}_{1}=\frac{\sqrt{3}{v}^{2}}{3g}$£¬BÍÑÀ뵯»Éºóʱ¼ätÄÚ·¢ÉúµÄλÒÆΪx₂=v₂t=2vt£¬
BÍÑÀ뵯»Éºóµ½BCÂäµØ£¬A¹²·¢ÉúµÄλÒÆΪ${x}_{3}={v}_{3}£¨t+{t}_{1}£©=vt+\frac{\sqrt{3}{v}^{2}}{3g}$£¬
ÔòBCÂäµØʱÓëPµãµÄˮƽ¾àÀëx=${x}_{1}+{x}_{2}+{x}_{3}=\frac{\sqrt{3}{v}^{2}}{g}$
´ð£º£¨1£©BCײºóÕûÌåµÄËÙ¶Èv₁Ϊv£»
£¨2£©µ¯»ÉѹËõºó¾ßÓеÄ×î´óµ¯ÐÔÊÆÄÜΪ3mv²£»
£¨3£©BCÂäµØʱÓëPµãµÄˮƽ¾àÀëxΪ$\frac{\sqrt{3}{v}^{2}}{g}$£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËƽÅ×Ô˶¯µÄ»ù±¾¹æÂÉ¡¢¶¯Á¿Êغ㶨ÂÉÒÔ¼°ÄÜÁ¿Êغ㶨ÂɵÄÖ±½ÓÓ¦Óã¬ÒªÇóͬѧÃÇÄÜÕýÈ··ÖÎöÎïÌåµÄÔ˶¯¹ý³ÌÖУ¬Ñ¡ÔñºÏÊʵĶ¨ÂÉÇó½â£¬ÖªµÀÔÚÅöײ¹ý³ÌÖУ¬¶¯Á¿Êغ㣬ÄѶÈÊÊÖУ®