Does the edge of the shadow of a person's head move in a straight line if the person moves in a straight line with the light source fixed in a place?

Question

So, let's say that there is a light source at a height "H" above the ground. A person of height "x" starts moving in a straight line with uniform velocity (not direct under the source of light) and passes the light source at a point. Now what I am dubious about is that, if the edge of the shadow of the person's head also move in a straight line during the motion too? I was initially presuming that it would, but then when I tried applying the Pythagorean theorem for the triangle (formed by connecting the edge of the shadow of the head to the source, the line connecting edge of the shadow to the wall of the source, and the source with the wall), it seems like the distance between the edge of the shadow from the source of light is different at each point. I really am not sure even though it seems like I vouch for the latter (not being in the straight line), but I wish to know and look forward to understand what really it is.

Answer 1

I am assuming the shadow of the top of the walker's head falls on the same level ground on which he is walking.

Suppose the light source is at $L$ vertically above point $O$ on the ground and $N$ is the point on the walker's path closest to $O$, where $ON=d$. At some time $t$ the walker of height $x$ has moved to a point $P$ so that the top of his head, point $X$, makes a shadow at point $Q$ on the ground.

Let the perpendicular distance from $Q$ to the path be $y$, and let $\angle NOP=\theta$. Then we have

$OP=d\csc\theta$ and $PQ=y\csc\theta$.

Then by similar triangles, $$\frac{H}{x}=\frac{(d+y)\csc \theta}{y\csc\theta}=\frac{d+y}{y}$$

Rearranging, $$y=\frac{xd}{H+x}$$

This indicates that $y$ remains constant and the path of the shadow is therefore a straight line parallel to the walker's. The walker doesn't have to move at constant speed.