跳伞运动员做低空跳伞表演,当直升飞机悬停在离地面224m高时,运动员离开飞机作自由落体运动.运动一段时间后,打开降落伞,
3个回答

哦 哦 哦

v1=gt=√ 2(H-h)

2ah=(v1)2-(v2)2=~

where v1 is the initial speed that moment the jumper open his

and v2 is the final speed H is the total distance and h is when he opened

h'=(v2)22g

第二问不明白最短是啥意思

T=t1+t2=√2(H-h)g+(v1-v2)g

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