请教physics高手

yangwei

A stone is drop from the top of building. 2s later, a second stone is throw downward with an initial speed of 3m  s^-2. Both stone landed on the ground at the same time.
a) how long it takes for the first stone to reach the ground
b) how high is the building
c) what r the speed of the stones just before they hit the ground

谢啦...

meiling

initial speed of 3m  s^-2
speed不是ms^-1的么？

我把那个3ms^-2当成3ms^-1
我认为：
for second stone
s=ut+0.5at^2
s=3t+0.5(9.81)t^2
for 1st stone
s=ut+0.5at^2   （initial speed is zero)
s=0.5(9.81)(t+2)^2
they hav common s
therefore
3t+0.5(9.81)t^2=0.5(9.81)(t+2)^2
然后你自己算。。

[ 本帖最后由 meiling 于 4-7-2009 06:27 PM 编辑 ]

yangwei

是我搞错了...ms^-1

答案是negative不用紧吗..

谢谢...

yangwei

是我搞错了...ms^-1

答案是negative不用紧吗..

谢谢...

yangwei

但是如果我用1st stone 跟2nd stone的time来做b。。答案差很远...不是应该一样吗...c的是找v吗

meiling

negative应该是因为那个a。。
a就要放-9.81。。
对，c)是要找v

咕叽咕叽

这个问题应该是错了。。两秒过后，第一个石头的velocity已经是19.62ms-1了，v=9.81x2=19.62。那时，第二个石头才以3ms-1的velocity掉下。由于acceleration是一样，第二个石头永远也追不上第一个石头，所以不可能同时落地。。