從物理學(xué)角度,需要用運(yùn)動學(xué)配合能量守恒解,完全用能量守恒是是得不出時間的.
從數(shù)學(xué)意義上,這是一道數(shù)學(xué)級數(shù)求極限的題.
t0=v0/g=根下(2h/g)
t1=v1/g=(3/4v0)/g=3/4t0
同理:
t2=3/4t1=(3/4)^2t0
.
tn=(3/4)^nt0
t=t0+2t1+2t2+.2tn
=t0+2t0[(3/4)+(3/4)^2+.(3/4)^n]
=t0+2t0*(3/4)*[1-(3/4)^n]/(1-3/4)
當(dāng)n趨近于+∞時
總時間：t=t+2t0*(3/4)*1/(1-3/4)=t0+6t0=7t0=7根下(2h/g)
mgh1=1/2mv1^2=1/2m（v0*3/4)^2=(3/4)^2mgh
h1=(3/4)^2h
s1=2h1=2*(3/4)^2h
同理:
h2=(3/4)^4h
h3=(3/4)^6h
.
hn=(3/4)^(2n)h
s=h+s1+s2+s3+.+sn
=h+2(h1+h2+H3+.hn)
=h+2h[(3/4)^(2*1)+(3/4)^(2*2)+.(3/4)^(2*n)]
=h+2h*(3/4)^2*{1-[(3/4)^2]^n}/[1-[(3/4)^2]
當(dāng)n趨近于+∞時,
總路程s=h+2h*(3/4)^2*1/[1-[(3/4)^2]=25h/7