推荐回答（3个）
哈哈，首先觉得这烟花很搞笑，只炸成2块。
思想：
1：第一个条件，给了你系统的总动量，以及下一个问题的线索。
2：从500米处，下降10秒。可以知道其中一半的初速度。
3：联立可以解决。
解答：
1：系统总动量：30*m（m为礼花弹的质量）
2：向下部分初速度：v2*v2-v1*v1=2*g*500
v2=v1+g*10
得：v1=0
向下部分动量：0*m/2=0
3: 30m=v*m/2+0
得 v=60m/s
解：设竖直下落的初速度为V1
由运动学公式：10*10*10/2+V1*10=500
得V1=0，即下落的做自由落体运动
设原礼花弹质量为M，由动量守恒得：
MV=1/2MV2 代入数据得：V2=60m/s 方向竖直向上
所以另一部分的速度为60m/s 方向竖直向上
查了一下书,我们用的是不相同课本!
这样的烟花很危险,不是搞笑!
先算出炸后的速度,上楼的已经用公式法,也可用平均速度!初学者多体会
500m 走10s,可见平均速度为50m/s,也是中间时刻的速度,也就是用5s以10/ms2加速度,加速到50m/s,可以初速为0
在竖直方向上,因为内力远大于外力,所以认为在竖直的方向上动量守恒!
MV=1/2MV2
算得60m/s
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