高一物理，高手进！
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v^2=2aL
(v/2)^2=2aL'
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1/4=L'/L
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L'=L/4
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mg(h+D)-fD=0
f=mg(h+D)/D
在泥中运动时，ma=f-mg
又 mgh=mv^2/2
t=v/a
得 t=D√(2/gh)
当我没看见。。。
mgh=w...1
w=FD...2
1,2得F=mgh/D
mgh=1/2m v^2...3
mv=Ft...4
3,4得t=（根号下2gh)D/gh
1、物体的势能全部用于克服淤泥对物体阻力做功。得
（f-mg）D=mg(h+D)
f=mg+mg（h+D）/D =mg(h+2D)/D
2、下落到地面时机械能守恒mgh=(1/2)mV^2, 得V=(2gh)^(1/2)
冲量定理（f-mg）t=mV=m(2gh)^(1/2)，
t=D*(2h/g)^(1/2)/(h+D)
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