Question

18£®ÔÚÒ»ÇãбµÄdzɫ³¤´«ËÍ´øµ×¶ËÇáÇá·ÅÖÃһú¿é£¨¿ÉÊÓΪÖʵ㣩£¬³¤´«ËÍ´øÓëˮƽÃæ¼Ð½ÇΪ¦È£¬³¤¶ÈΪL£¬Ãº¿éÓë´«ËÍ´øÖ®¼äµÄ¶¯Ä¦²ÁÒòÊýΪ¦Ì£®³õʼʱ£¬´«ËÍ´øÓëú¿é¶¼ÊǾ²Ö¹µÄ£®ÏÖÈô«ËÍ´øÒԺ㶨µÄ¼ÓËÙ¶Èa₀¿ªÊ¼Ô˶¯£¬µ±ÆäËٶȴﵽv₀ºó£¬±ãÒÔ´ËËÙ¶È×öÔÈËÙÔ˶¯£®¾­¹ýÒ»¶Îʱ¼ä£¬Ãº¿éÔÚ´«ËÍ´øÉÏÁôÏÂÁËÒ»¶ÎºÚÉ«ºÛ¼£ºó£¬Ãº¿éÏà¶ÔÓÚ´«ËÍ´ø²»ÔÙ»¬¶¯£®£¨×î´ó¾²Ä¦²ÁÁ¦¿ÉÊÓΪ»¬¶¯Ä¦²ÁÁ¦£©
£¨1£©Çó´ËºÚÉ«ºÛ¼£µÄ³¤¶È£»
£¨2£©Çóú¿éÔËË͵½¶¥¶ËËùÐèʱ¼ä£®

Answer 1

·ÖÎö £¨1£©´«ËÍ´øÏȼÓËÙºóÔÈËÙ£¬Ãº¿éÏȼÓËÙ£¬µ±ËÙ¶ÈÔö´óµ½µÈÓÚ´«ËÍ´øËٶȺó£¬Óë´«ËÍ´ø²»ÔÙÏà¶Ô»¬¶¯£¬Ò»ÆðÔÈËÙ£®Ïȸù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³ö¼ÓËٶȣ¬ÔÙ¸ù¾ÝÔ˶¯Ñ§¹«Ê½¼ÆËã³öÔ˶¯Ê±¼ä£¬Çó³öú¿éºÍƤ´øµÄλÒƺ󣬼´¿ÉÒԵõ½ºÚÉ«ºÛ¼£µÄ³¤¶È£®
£¨2£©Ãº¿éÔËË͵½¶¥¶ËµÄʱ¼äÊÇÔȼÓËÙʱ¼äºÍÔÈËÙÔ˶¯Ê±¼äÖ®ºÍ

½â´ð ½â£º£¨1£©Ãº¿éÊÜÁ¦·ÖÎö£¬Êܵ½ÖØÁ¦¡¢Ö§³ÖÁ¦ºÍÏòÉϵĻ¬¶¯Ä¦²ÁÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉ£¬ÓÐ
¦Ìmgcos¦È-mgsin¦È=ma
½âµÃ£ºa=¦Ìgcos¦È-gsin¦È
ÔȼÓËÙʱ¼ä$t=\frac{{V}_{0}^{\;}}{a}=\frac{{V}_{0}^{\;}}{¦Ìgcos¦È-gsin¦È}$
ú¿éλÒÆ£º${x}_{1}^{\;}=\frac{{V}_{0}^{\;}}{2}t$
´«ËÍ´øλÒÆ£º${x}_{2}^{\;}=\frac{{V}_{0}^{2}}{2{a}_{0}^{\;}}$
Ïà¶ÔλÒÆ£º$¡÷x={x}_{1}^{\;}-{x}_{2}^{\;}=\frac{{V}_{0}^{\;}}{2}t-\frac{{V}_{0}^{2}}{2{a}_{0}^{\;}}$=$\frac{{V}_{0}^{2}}{2}£¨\frac{1}{g£¨¦Ìcos¦È-sin¦È£©}-\frac{1}{{a}_{0}^{\;}}£©$
ºÚÉ«ºÛ¼£µÄ³¤¶ÈµÈÓÚú¿éÓë´«ËÍ´ø¼äµÄÏà¶ÔλÒÆ£»
£¨2£©Ãº¿éÔȼÓËÙÔ˶¯µÄʱ¼ä${t}_{1}^{\;}=\frac{{V}_{0}^{\;}}{a}$
ú¿éÔÈËÙÔ˶¯µÄλÒÆ£º$L-\frac{{V}_{0}^{2}}{2a}$
ÔÈËÙÔ˶¯µÄʱ¼ä£º${t}_{2}^{\;}=\frac{L-\frac{{V}_{0}^{2}}{2a}}{{V}_{0}^{\;}}=\frac{L}{{V}_{0}^{\;}}-\frac{{V}_{0}^{\;}}{2a}$
×Üʱ¼ä£º$t={t}_{1}^{\;}+{t}_{2}^{\;}=\frac{L}{{V}_{0}^{\;}}+\frac{{V}_{0}^{\;}}{2a}$=$\frac{L}{{V}_{0}^{\;}}+\frac{{V}_{0}^{\;}}{2g£¨¦Ìcos¦È-sin¦È£©}$
´ð£º£¨1£©´ËºÚÉ«ºÛ¼£µÄ³¤¶È$\frac{{V}_{0}^{2}}{2}£¨\frac{1}{g£¨¦Ìcos¦È-sin¦È£©}-\frac{1}{{a}_{0}^{\;}}£©$£»
£¨2£©Ãº¿éÔËË͵½¶¥¶ËËùÐèʱ¼ä$\frac{L}{{V}_{0}^{\;}}+\frac{{V}_{0}^{\;}}{2g£¨¦Ìcos¦È-sin¦È£©}$

µãÆÀ ±¾Ìâ¹Ø¼ü·ÖÎöÇå³þƤ´øºÍú¿éµÄÊÜÁ¦Çé¿öºÍÔ˶¯Çé¿ö£¬È»ºó¸ù¾ÝÔ˶¯Ñ§¹«Ê½Çó½â£®