Ladder

4 m long ladder touches the cube 1mx1m at the wall. How high reach on the wall?

Correct result:

q₁ =  1.3622 m
q₂ =  3.7609 m

Solution:

$$\begin{gathered} l=4 \\ f(x)=k_x+q \\ f(1)=1 \\ f(x_0)=0 \\ k=1\mathrm{/}(1-x_0) \\ q=1-k \\ {x}_{01}=3.7609 \\ {x}_{02}=1.3622 \\ {k}_1=1\mathrm{/}(1-x_{01})=1\mathrm{/}(1-3.7609)\doteq -0.3622 \\ {q}_1=1-k_1=1-(-0.3622)\doteq 1.3622=1.3622\text{ m } \\ {l}_1=\sqrt{q_1^2+x_{01}^2}=\sqrt{1.3622^2+3.7609^2}\doteq 4 \end{gathered}$$

$$\begin{gathered} k_2=1\mathrm{/}(1-x_{02})=1\mathrm{/}(1-1.3622)\doteq -2.7609 \\ {q}_2=1-k_2=1-(-2.7609)\doteq 3.7609=3.7609\text{ m } \\ {l}_2=\sqrt{q_2^2+x_{02}^2}=\sqrt{3.7609^2+1.3622^2}=4 \end{gathered}$$

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