Not always. My expertly drawn paint picture explains. The red segment is the set of possible lighthouse locations, the black areas are the possible closest lands from lighthouses, the blue is the perimeter and the green is just there to depict circles I drew. For the ship to definitely find land, it needs to cruise along the blue perimeter since it is possible that everything inside it is water. If the radius is 10km, the length of the perimeter is 2r*pi+2r or around 82km and accounting the distance to get to this perimeter in the first place this is far more than 75km (although for such a loose inequality my solution is probably false).

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The idea is same as above but I think you need to formalize it a bit. Consider the below blue shape $B$ with boundary $\Gamma$ (convex hull of a circle of radious $10$ km centered at the ship $S$ and the furthest possible lighthouse $L$). Considering the brown waterside as an example, it's easy to see a line can be a waterside iff it intersects $\Gamma$ at atlast a point (or touches it). [asy][asy] pair O = (0,0), A = (1,0), L = (0.3,0); path b = arc(O, 1, 90,270)--(1,-1)--arc(A,1,-90,90)--cycle; path c = circle(L,1); filldraw(b,paleblue,deepblue); dot(O^^A^^L); draw(O--A, red); draw(c, red+dotted); pair tan = dir(c, .2); path tangent = shift(point(c,.2)) * scale(1.2)* (-tan -- tan); draw (tangent, brown); label("$S$", O, align = W); label("$L$", A, align = E); [/asy][/asy] Lemma Let $C_1, C_2$ be two closed convex sets such that every line intersecting/touching the boundary of $C_1$ also intersects the boundary of $C_2$. Then $C_1$ is completely contained in $C_2$. Proof: Assume not. Then pick a point $P \in C_1 - (C_1 \cap C_2)$. Since $C_1, C_2$ are closed we can find an open ball $B$ centered at $P$ completely contained in $C_1$ and not intersecting $C_2$. Let $\omega$ be a circle centered at $P$ contained inside $B$, and color a point $Q \in \omega$ red if $\overrightarrow{PQ} \cap C_2$ is nonempty. Now note that if all red points don't lie in a semicircle, then there exists three red points $Q_1, Q_2, Q_3 \in \omega$ forming an acute triangle are three points $A_1, A_2, A_3 \in C_2$ such that $PQ_iA_i$ are colinear in that order. Since an acute angled traingle has circumcenter inside it, $P \in \Delta Q_1Q_2Q_3$, and thus by convexity, $ P \in \Delta Q_1Q_2Q_3 \in \Delta A_1A_2A_3 \subset C_2 $ a contradiction ! So suppose all red points are in the semicircle $B_1PB_2$ (with $B_1, B_2 \in \omega$). [asy][asy] path p1 = scale(10)*((1,0)..(1,.5)..(0,1.2)..(-1,1)..(-1,-1)..cycle); path p2 = scale(10)*((.9,0)..(.4,.5)..(-1.5,1.2)..(-1,-.6)..(.8,-1.2)..cycle); draw(p1, green); draw(p2, blue); label(Label("$C_1$", position=Relative(0.2), align = E), p1); label(Label("$C_2$", position=Relative(0.6), align = W), p2); pair P = scale(10)*(0.6, 0.65); dot(P); path w1 = Circle(P, 2); draw(w1); draw(arc(P, 2, 140, 310), red); path l = shift(P)*scale(15)*(dir(135)--dir(315)); draw(l, L = Label("$\ell$", position = EndPoint)); [/asy][/asy] Now take any line $\ell$ parallel to $B_1B_2$ intersecting $\omega$ in the side of the semicircle not containing any red points. I claim that $C_2$ don't intersect $\ell$. Indeed, if $A \in C_2 \cap \ell$ then $\overrightarrow{PA} \cap \omega$ would be red point not in the intended semicirlce, a contradiction ! $\blacksquare$ Now, let the proposed tranjectory be $\gamma$ such that every line intersecting $\Gamma$ also intersects $\Gamma$ . If $\gamma$ isn't convex, then we can take the "convex hull" of it without decreasing it's perimeter. By the above Lemma, $\Gamma$ is completely contained inside $\gamma$, the perimeter of $\gamma$ is not less than the perimeter of $\Gamma = 2r(\pi +1) = 86.28$ km$, which implies the captain can't land (I'm also sceptical of this proof since the bound looks bit too weak)

The answer to this problem is actually yes - it is possible to find such a way (you can see the official solution for more details). The solution above is very interesting and well-written, but the mistake is that actually the trajectory of the ship isn't necessarily a closed convex set and therefore we cannot apply the lemma. In fact, it is enough for us to find a (perhaps open) set $\gamma$, so that every line that intersects $\Gamma$ also intersects $\gamma$ and this turns out to be possible. So the problem is actually a bit easier than the above solution, but I still think that the above is edifying and the results in it may be very useful for solving other problems.