A girl of height 100 cm …

Let AB be the lamp-post, DE be the girl and D be the position of girl after 4s.
Again, let DC = xm be the length of shadow of the girl.
Given, DE = 100 cm =1 m, AB =5 m and speed of the girl= 1.9 m/s
{tex}\therefore{/tex} Distance of the girl from lamp-post after 4 s = BD = {tex}1.9\times4=7.6\;m{/tex} [{tex}\;\because{/tex} Distance = speed {tex}\times{/tex} time ]
In {tex}\triangle ABC{/tex} and {tex}\triangle EDC{/tex},  {tex}\angle B=\angle D{/tex}
[{tex}\;\because{/tex}each {tex}90^o{/tex}] 
{tex}\angle C=\angle C{/tex} [ {tex}\;\because{/tex}common angle ] 
{tex}\therefore \triangle ABC\sim \triangle EDC{/tex} [{tex}\;\because{/tex} by AA similarity criterion ]
{tex}\Rightarrow \frac{BC}{DC}=\frac{AB}{DE}{/tex} {tex}\;\because{/tex}[since, corresponding sides of similar triangles are proportional]...(i)
On substituting all the values in Eq(i), we get 
{tex}\frac{7.6+x}{x}=\frac{5}{1}{/tex} 
{tex}\Rightarrow 7.6+x=5x{/tex}
{tex}\Rightarrow 7.6 =5x- x{/tex}
{tex}\Rightarrow 7.6 =4x{/tex}
{tex}\Rightarrow x =\frac{7.6}{4}{/tex} 
{tex}\Rightarrow x=1.9\;m{/tex}
Hence the length of her shadow after 4 s is 1.9 m.