Question

17£® ÈçͼËùʾ£¬ÔÚͶÇòÓÎÏ·ÖУ¬Ð¡Ã÷×øÔÚ¿ÉÑØÊúÖ±·½ÏòÉý½µµÄÒÎ×ÓÉÏ£¬ÏòÕýÇ°·½·ÅÔÚˮƽµØÃæÉϵÄԲͰÖÐÅ×СÇò£¨¿ÉÊÓΪÖʵ㣩£®Ä³´ÎÔڸ߶ÈH₀´¦½«Ð¡ÇòÒÔËÙ¶Èv₀ˮƽÅ׳ö£¬Ð¡ÇòÂäÈëÍ°ÖÐÇÒδ´¥ÅöÍ°±Ú£®ÒÑ֪ͰµÄÖ±¾¶Îªd£¬¸ß¶ÈΪh₀£¬Å׳öµãÓëÍ°±ÚµÄˮƽ¾àÀëʼÖÕΪL£¬ºöÂÔ¿ÕÆø×èÁ¦£®
£¨1£©ÇóСÇò¸ÕÂ䵽Ͱµ×¾­ÀúµÄʱ¼ä£»
£¨2£©ÇóСÇò¸ÕÂ䵽Ͱµ×ʱµÄËٶȴóС£»
£¨3£©ÈôСÇòÒÔ²»Í¬ËÙ¶ÈˮƽÅ׳öºó¶¼ÄÜÂäÈëÍ°ÖУ¬ÇÒÈëͰʱ±£Ö¤²»´¥ÅöÍ°±Ú£¬Çó¸ÃËٶȵķ¶Î§£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝϽµµÄ¸ß¶ÈÇó³öСÇò¸ÕÂ䵽Ͱµ×¾­ÀúµÄʱ¼ä£»
£¨2£©¸ù¾ÝËÙ¶Èʱ¼ä¹«Ê½Çó³öСÇòµ½´ïÍ°µ×µÄÊúÖ±·ÖËٶȣ¬½áºÏƽÐÐËıßÐζ¨ÔòÇó³öСÇòµ½´ïÍ°µ×µÄËٶȴóС£®
£¨3£©¸ù¾ÝСÇòÂäÔÚÇ°ÑغͺóÑØˮƽλÒÆ£¬½áºÏÔ˶¯µÄʱ¼äÇó³öËٶȵķ¶Î§£®

½â´ð ½â£º£¨1£©¸ù¾Ý${H}_{0}=\frac{1}{2}g{t}^{2}$µÃ£¬Ð¡Çò¸ÕÂ䵽Ͱµ×¾­ÀúµÄʱ¼ät=$\sqrt{\frac{2{H}_{0}}{g}}$£®
£¨2£©Ð¡Çòµ½´ïÍ°µ×µÄÊúÖ±·ÖËÙ¶È${v}_{y}=\sqrt{2g{H}_{0}}$£¬
¸ù¾ÝƽÐÐËıßÐζ¨ÔòÖª£¬Ð¡Çò¸ÕÂ䵽Ͱµ×µÄËٶȴóСv=$\sqrt{{{v}_{0}}^{2}+{{v}_{y}}^{2}}$=$\sqrt{{{v}_{0}}^{2}+2g{H}_{0}}$£®
£¨3£©¸ù¾Ý${H}_{0}-{h}_{0}=\frac{1}{2}g{{t}_{1}}^{2}$µÃ£¬${t}_{1}=\sqrt{\frac{2£¨{H}_{0}-{h}_{0}£©}{g}}$£¬ÔòËٶȵÄ×îСֵ${v}_{1}=\frac{L}{{t}_{1}}=L\sqrt{\frac{g}{2£¨{H}_{0}-{h}_{0}£©}}$£®
µ±Ð¡ÇòÂäÈëÍ°ÖÐˮƽ¾àÀë¸ÕºÃµÈÓÚL+dʱ£¬ËÙ¶È×î´ó
ËٶȵÄ×î´óÖµv₂=$\frac{L+d}{t}$=$\frac{L+d}{\sqrt{\frac{2{H}_{0}}{g}}}$=£¨L+d£©$\sqrt{\frac{g}{2{H}_{0}}}$£®
Ëٶȵķ¶Î§ÎªL$\sqrt{\frac{g}{2£¨{H}_{0}-{h}_{0}£©}}$£¼v£¼£¨L+d£©$\sqrt{\frac{g}{2{H}_{0}}}$£®
´ð£º£¨1£©Ð¡Çò¸ÕÂ䵽Ͱµ×¾­ÀúµÄʱ¼äΪ$\sqrt{\frac{2{H}_{0}}{g}}$£®
£¨2£©Ð¡Çò¸ÕÂ䵽Ͱµ×ʱµÄËٶȴóСΪ$\sqrt{{{v}_{0}}^{2}+2g{H}_{0}}$£®
£¨3£©Ëٶȵķ¶Î§ÎªL$\sqrt{\frac{g}{2£¨{H}_{0}-{h}_{0}£©}}$£¼v£¼£¨L+d£©$\sqrt{\frac{g}{2{H}_{0}}}$£®

µãÆÀ ½â¾ö±¾ÌâµÄ¹Ø¼üÖªµÀƽÅ×Ô˶¯ÔÚˮƽ·½ÏòºÍÊúÖ±·½ÏòÉϵÄÔ˶¯¹æÂÉ£¬×¥×¡ÁÙ½ç״̬£¬½áºÏÔ˶¯Ñ§¹«Ê½Áé»îÇó½â£®