五个直径均为d=5 cm的圆环连接在一起,用细线悬于O点.枪管水平时

五个直径均为d=5 cm的圆环连接在一起,用细线悬于O点.枪管水平时枪口中心与第五个环的环心在同一水平面上,如图5-3-15所示.它们相距100 m,且连线与球面垂直.现烧断细线,经过0.1 s后开枪射出子弹,若子弹恰好穿过第2个环的环心,求子弹离开枪口时的速度(不计空气阻力,g10 m/s2).

5-3-15

答案

解析: 设从子弹射出到穿过环心所用时间为t,则根据平抛运动在竖直方向上做自由落体运动的特点,得竖直方向的位移关系:

s+0.05×(5-2)m=s

解得t=0.1 s.

又据子弹水平方向做匀速直线运动,则

m/s=1 000 m/s.

答案:1 000 m/s

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