Step 1: Recall what decides the range when the speed is fixed.
In projectile motion the horizontal range for a given launch speed $u$ is
\[ R = \frac{u^2 \sin 2\theta}{g} \]
All four athletes run at the same speed $u = 30$ kmph, so the range depends only on $\sin 2\theta$. Whoever jumps at the angle that makes $\sin 2\theta$ largest wins, without needing a separate calculation of $u$.
Step 2: Use the symmetry of the sine function instead of computing every value from scratch.
$\sin 2\theta$ reaches its single largest value, $1$, exactly when $2\theta = 90^{\circ}$, that is when $\theta = 45^{\circ}$. Moving further away from $45^{\circ}$ on either side makes $\sin 2\theta$ smaller, because sine is symmetric about $90^{\circ}$: $\sin(90^{\circ}+x) = \sin(90^{\circ}-x)$.
Step 3: Compare each athlete's angle to $45^{\circ}$.
A is at $30^{\circ}$, which is $15^{\circ}$ away from $45^{\circ}$. B is at $45^{\circ}$ exactly, $0^{\circ}$ away. C is at $60^{\circ}$, also $15^{\circ}$ away, so A and C land at the same range by that symmetry ($\sin 60^{\circ}=\sin 120^{\circ}$). D is at $75^{\circ}$, which is $30^{\circ}$ away, the furthest of all, so D jumps the shortest distance.
Step 4: Pick the winner.
Since B's angle is exactly $45^{\circ}$, B has zero deviation from the optimum angle and therefore the largest possible $\sin 2\theta$, equal to $1$, giving the maximum range among the four.
Final Answer:
The competition is won by the athlete jumping closest to the optimum $45^{\circ}$ angle.
\[ \boxed{\text{B}} \]