Question

15£®ÈçͼËùʾ£¬ÔÚÖ±½Ç×ø±êxOyÄÚ£¬ÔÚµÚÒ»ÏóÏÞµÄÇøÓòIÄÚ´æÔÚ´¹Ö±ÓÚÖ½ÃæÏòÍâ¿í¶ÈΪdµÄÔÈÇ¿´Å³¡£¬ÇøÓò¢òÄÚ´æÔÚ´¹Ö±ÓÚÖ½ÃæÏòÀï¿í¶ÈΪ$\frac{d}{3}$µÄÔÈÇ¿´Å³¡£»ÔÚµÚÈýÏóÏÞ´æÔÚÑØyÖáÕýÏòµÄÔÈÇ¿µç³¡£¬Ò»ÖÊÁ¿Îªm¡¢´øµçÁ¿Îª+qµÄ´øµçÁ£×Ӵӵ糡ÖеÄ×ø±êΪ£¨-2h£¬-$\sqrt{3}$h£©µãÒÔËÙ¶Èv₀ˮƽÏòÓÒÉä³ö£¬¾­¹ýÔ­µãO´¦ÉäÈëÇøÓò¢ñºó´¹Ö±MNÉäÈëÇøÓò¢ò£¬£¨Á£×ÓµÄÖØÁ¦ºöÂÔ²»¼Æ£©Çó£º
£¨1£©ÇøÓò¢ñÄڴŸÐӦǿ¶ÈB₁µÄ´óС£»
£¨2£©ÈôÇøÓò¢òÄڴŸÐӦǿ¶ÈB₂µÄ´óСÊÇB₁µÄÕûÊý±¶£¬µ±Á£×ÓÔٴλص½MNʱ×ø±ê¿ÉÄÜֵΪ¶àÉÙ£¿

Answer 1

·ÖÎö £¨1£©Á£×ÓÔڵ糡ÖÐ×öÀàƽÅ×Ô˶¯£¬Ôڴų¡ÖÐ×öÔÈËÙÔ²ÖÜÔ˶¯£¬ÓÉƽÅ×Ô˶¯µÄ¹æÂÉ¿ÉÇóµÃ½øÈë´Å³¡µÄËٶȣ¬ÔÙÓÉÔ²ÖÜÔ˶¯µÄ¹æÂÉÁгö·½³Ì£¬ÁªÁ¢¿É½âµÃB₁£»
£¨2£©Á£×ÓÔÚÁ½´Å³¡¾ù×öÔÈËÙÔ²ÖÜÔ˶¯£¬Óɼ¸ºÎ¹Øϵ¿ÉÖªÁ£×ÓÔÚB2ÖеÄת¶¯°ë¾¶£¬¿É·ÖÎöÁ£×ÓÄÜ·ñ»Øµ½MN£¬ÔÙ·Ö±ð½øÐÐÌÖÂÛ¼´¿ÉµÃ³öÕýÈ·½áÂÛ£®

½â´ð ½â£º£¨1£©´øµçÁ£×ÓÔÚÔÈÇ¿µç³¡ÖÐ×öÀàƽÅ×Ô˶¯£¬Éèµ½´ïÔ­µãʱÊúÖ±·½ÏòËÙ¶ÈΪv_y£¬
¿ÉµÃ 2h=v₀t  
$\sqrt{3}h=\frac{{v}_{y}}{2}t$£¬
¿ÉµÃ$\sqrt{3}$v₀=v_y
ÓÉÔ˶¯µÄºÏ³É¿ÉµÃv=2v₀£¬·½ÏòÓëxÖáÕýÏò³É60¡ã£®
Á£×Ó½øÈëÇøÓòI×öÔÈËÙÔ²ÖÜÔ˶¯£¬Óɼ¸ºÎ֪ʶ¿ÉµÃ
R₁=$\frac{d}{sin60¡ã}$=$\frac{2\sqrt{3}d}{3}$£®
ÓÉÂåÂ××ÈÁ¦³äµ±ÏòÐÄÁ¦ B₁qv=m$\frac{{v}^{2}}{{R}_{1}}$£¬
¿É½âµÃB₁=$\frac{\sqrt{3}m{v}_{0}}{qd}$£®
£¨2£©Éè B₂=nB₁£¬Ôò R₂=$\frac{{R}_{1}}{n}=\frac{2\sqrt{3}d}{3n}$£®
²»´ÓB₂ÓÒ²à·É³ö£¬ÓРR₂£¼$\frac{d}{3}$£¬
¼´$\frac{2\sqrt{3}d}{3n}£¼\frac{d}{3}$£¬£¨n$£¾2\sqrt{3}$£¬È¡n=4£¬5£¬6£¬¡­£©  
Á£×Ó½øÈëB₂´¦µÄ×Ý×ø±ê${y}_{1}={R}_{1}-\sqrt{{{R}_{1}}^{2}-{d}^{2}}=\frac{\sqrt{3}}{3}d$£¬
ÔÚB₂ÖÐת°ëÔ²×Ý×ø±ê${y}_{2}=\frac{{R}_{1}}{2}+2{R}_{2}=\frac{\sqrt{3}d}{3}+\frac{4\sqrt{3}d}{3n}$ £¨È¡n=4£¬5£¬6£¬¡­£©
ºá×ø±ê x₂=d£®
´ð£º£¨1£©ÇøÓò¢ñÄڴŸÐӦǿ¶ÈB₁µÄ´óСΪ$\frac{\sqrt{3}m{v}_{0}}{qd}$£®
£¨2£©Á£×ÓÔٴλص½MNʱ×ø±ê¿ÉÄÜֵΪ£¨d£¬$\frac{\sqrt{3}d}{3}+\frac{4\sqrt{3}d}{3n}$£©£¬£¨È¡n=4£¬5£¬6£¬¡­£©

µãÆÀ ´øµçÁ£×ÓÔڴų¡ÖеÄת¶¯£¬Òª×¢ÒâÈ·¶¨Ô²ÐĺͰ뾶£¬ÕÒ³öÕýÈ·µÄ¼¸ºÎ¹Øϵ£»±ØҪʱҪעÒâÌÖÂÛ£¬ÕÒ³öËùÓеĿÉÄÜÇé¿öÔÙ×ö³ö½áÂÛ£®