淑儀站在泳池的彈板上(圖q)。她跳到半空，到達最高點後再向下落，然後進入水裹。(假設室氣阻力忽略不計。)
(a)淑儀跳起後0.7s到達最高黠。她離問彈板時的速率是多少?
(b)她從最高點下落至接觸水面需時1.5s。彈板與水面的距離是多少？
物理力學問題(15分)
2008-11-04 4:43 am
回答 (4)
2008-11-08 5:07 am
圖Q呢?最高點在那?
無圖幫唔到你~
無圖幫唔到你~
2008-11-05 4:03 am
a) 0.7s到達最高黠
即係表示０．７ｓ時佢的速率係０．
ａ＝－１０
所以 利用v=u+at
0＝u＋(-１０)*０．７
u=7
a=10 u=7 t=(1.5-0.7)=0.8
s=ut+at^2/2
s=0.8*7+10*0.8*0.8/2
s=5.6+3.2
s=8.8
即係表示０．７ｓ時佢的速率係０．
ａ＝－１０
所以 利用v=u+at
0＝u＋(-１０)*０．７
u=7
a=10 u=7 t=(1.5-0.7)=0.8
s=ut+at^2/2
s=0.8*7+10*0.8*0.8/2
s=5.6+3.2
s=8.8
參考: 我
2008-11-04 8:17 am
let me show you
(A).v=0 a=-10 t=0.7 u
(v-u)/t=a
u=7m/s
(b).the distance between the board and the highest point:
s=ut+0.5at^2
s=4.9+(-2.45)
=2.45m
the distance between the highest point and water surface:
s=ut+0.5at^2 (at the highest point u=0)
=0.5x10x1.5^2
=11.25m
so,the distance between the board and water surface is 11.25-2.45=8.8m
(A).v=0 a=-10 t=0.7 u
(v-u)/t=a
u=7m/s
(b).the distance between the board and the highest point:
s=ut+0.5at^2
s=4.9+(-2.45)
=2.45m
the distance between the highest point and water surface:
s=ut+0.5at^2 (at the highest point u=0)
=0.5x10x1.5^2
=11.25m
so,the distance between the board and water surface is 11.25-2.45=8.8m
參考: myself
2008-11-04 4:52 am
a.
v=u+at
0=u+10(0.7)
u=7ms^-1(向上)
b.
s=ut+0.5at^2
s=5.6+3.2
s=8.8m
v=u+at
0=u+10(0.7)
u=7ms^-1(向上)
b.
s=ut+0.5at^2
s=5.6+3.2
s=8.8m
參考: 自己
收錄日期: 2021-04-13 16:13:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081103000051KK01831