英语翻译After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?

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英语翻译After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?

英语翻译After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?
英语翻译
After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?

英语翻译After falling from rest at a height of 29 m,a 0.48 kg ball rebounds upward,reaching a height of 19 m.If the contact between ball and ground lasted 2.5 ms,what average force was exerted on the ball?
一个质量为0.48kg的球从29米位置自由下落(初速为0),下落后与地面接触时间为2.5ms,然后反弹至高度为19米处.问:此球与地面接触时地面平均给球施加多大的压力.
首先根据能量守恒定律算出球从29米处下落的地面的速度v1,和球在地面反弹时的速度-v2.(设竖直向下为正方向);
即:1/2mv^2=mgh,
v=(2gh)^(1/2)
求得v1=(2*10*29)^(1/2)=24.1m/s;
v2=(2*10*19)^(1/2)=19.5m/s.(g取10m/s*s)
再有动量守恒定律知:
冲量I=ft=mv1-(-mv2)=m(v1+v2)=0.48*(24.1+19.5)=21.0
则平均压力f=[m(v1+v2)]/t=21.0/0.0025=8371.2N

翻译
下降后从其余在高度29米,一0.48千克球篮板上升,达到高度19米如果球之间的接触和地面持续2.5毫秒,平均什么武力施加球?

翻译:
一个重0.48kg的球从高为29米的台子上落下,反弹后达到19米.如果球与地面接触时间为2.5毫秒,则球受到的平均力为多少?
求解:
从h1=29米落到地面时的速度v1为:
v1=根号(2gh1)=根号(2*29*10)=24.1m/s
反弹19米高,则地面处的反弹初速度v2为:
v1=根号(2gh2)=根号(2*19*10)=1...

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翻译:
一个重0.48kg的球从高为29米的台子上落下,反弹后达到19米.如果球与地面接触时间为2.5毫秒,则球受到的平均力为多少?
求解:
从h1=29米落到地面时的速度v1为:
v1=根号(2gh1)=根号(2*29*10)=24.1m/s
反弹19米高,则地面处的反弹初速度v2为:
v1=根号(2gh2)=根号(2*19*10)=19.5m/s
球与地面接触时动量守衡,接触时间t=2.5ms=2.5*10^(-3)s
设正方向为垂直向上,则v1=-24.1,v2=19.5
Ft=mv2-mv1
F=m(v2-v1)/t=0.48*[19.5-(-24.1)]/[2.5*10^(-3)]=8371(N)

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