以速度v0=20m/s竖直上抛一小球，2s后以相同的初速度在同一点竖直_爱下问搜题

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问题选择题

以速度v₀=20m/s竖直上抛一小球，2s后以相同的初速度在同一点竖直上抛另一小球，问两球在离抛出点多高处相遇（g取10m/s²）（　　）

A．10m处

B．15m处

C．20m处

D．不会相遇

答案

先判断先竖直上抛的小球到达最高点用的时间为：t=

v0

g

=

20

10

s═2s，

所以另一小球抛出时，它恰好在最高点将要做自由落体运动．由竖直上抛运动的对称性可得，两球再经过1s后相遇．

故两球相碰处离出发点的高度为：h=v0t-

1

2

gt²=20×1-

1

2

×10×12m=15m

故选：B

填空题

看图片，选词填空。

A. family B. six C. parents D. three
E. mother F. baby brother G. sister H. photo

1. Look! This is my family ____________. There are ____________ people in my family. They are my ____________ and me.
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单项选择题

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