Question

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Answer 1

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$mgsin37¡ã+{F}_{µç}^{\;}cos37¡ã=m{a}_{1}^{\;}$
$mgsin37¡ã+\frac{4mg}{3}¡Á\frac{4}{5}=m{a}_{1}^{\;}$
½âµÃ£º${a}_{1}^{\;}=\frac{5}{3}g$
ÉèBC=h
$AB=\frac{h}{sin37¡ã}$
ÓÉÔ˶¯Ñ§¹«Ê½£º$\frac{h}{sin37¡ã}=\frac{1}{2}{a}_{1}^{\;}{t}_{1}^{2}$
ÁªÁ¢½âµÃ${t}_{1}^{\;}=\sqrt{\frac{2h}{g}}$
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¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉ£º$mg=m{a}_{2}^{\;}$
½âµÃ£º${a}_{2}^{\;}=g$
ÓÉÔ˶¯Ñ§¹«Ê½£º$h=\frac{1}{2}{a}_{2}^{\;}{t}_{2}^{2}$
½âµÃ£º${t}_{2}^{\;}=\sqrt{\frac{2h}{g}}$
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¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉ£º${F}_{µç}^{\;}=\frac{4}{3}mg=m{a}_{3}^{\;}$
½âµÃ${a}_{3}^{\;}=\frac{4}{3}g$
¸ù¾Ý¼¸ºÎ¹Øϵ$BC=\frac{h}{tan37¡ã}$
¸ù¾ÝÔ˶¯Ñ§¹«Ê½£º$\frac{h}{tan37¡ã}=\frac{1}{2}{a}_{3}^{\;}{t}_{3}^{2}$
½âµÃ£º${t}_{3}^{\;}=\sqrt{\frac{2h}{g}}$
ËùÒÔ${t}_{1}^{\;}={t}_{2}^{\;}={t}_{3}^{\;}$
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