Point $A$ lies in the plane $\alpha$, the orthogonal projection of segment $A B$ onto this plane is $1, A B=2$. Find the distance from point $B$ to the plane $\alpha$.
Let $B 1$ be the orthogonal projection of point $B$ onto the plane $\alpha$. Then $B B 1$ is perpendicular to the plane $\alpha$, $A B 1$ is the orthogonal projection of segment $A B$ onto the plane $\alpha$, and the distance from point $B$ to the plane $\alpha$ is equal to the length of segment $B B 1$. The line $B B 1$ is perpendicular to the plane $\alpha$, so triangle $A B B 1$ is a right triangle. By the Pythagorean theorem

$$
B B 1=\sqrt{A B^{2}-A B^{2}}=\sqrt{4-1}=\sqrt{3} .
$$

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