Step 1: Calculate the horizontal range of the kicked football.
The range formula for projectile motion is $R = u^2 \sin 2\theta / g$. Here, $u = 30 \, \text{ms}^{-1}$, $\theta = 30^\circ$, and $g = 10 \, \text{ms}^{-2}$. So: \[ R = \frac{30^2 \sin 60^\circ}{10} = \frac{900 \times \frac{\sqrt{3}}{2}}{10} = 45\sqrt{3} \, \text{m} \] This is the total horizontal distance the ball covers from the kick point to where it lands.
Step 2: Find how far the second player must run.
The second player starts at $21\sqrt{3}$ m from the kick point and must reach the landing spot at $45\sqrt{3}$ m. The distance the second player must cover is: \[ d = 45\sqrt{3} - 21\sqrt{3} = 24\sqrt{3} \, \text{m} \]
Step 3: Calculate the total time of flight of the ball.
The time of flight is $T = 2u\sin\theta / g$: \[ T = \frac{2 \times 30 \times \sin 30^\circ}{10} = \frac{2 \times 30 \times \frac{1}{2}}{10} = \frac{30}{10} = 3 \, \text{s} \] The second player has exactly 3 seconds to reach the landing spot.
Step 4: Determine the minimum speed of the second player.
For the second player to just barely catch the ball, they must cover the distance $24\sqrt{3}$ m in exactly 3 seconds (reaching the landing point at the same time the ball arrives). Minimum speed: \[ v_{\min} = \frac{d}{T} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \, \text{ms}^{-1} \]
Step 5: Understand why this is the minimum speed.
If the second player runs faster, they arrive at the landing spot before the ball does and can still catch it. The minimum speed is defined as the speed required to just reach the final landing point exactly when the ball arrives, with no margin to spare. Any slower and the player cannot catch the ball at all.
Step 6: Verify the numerical answer.
$8\sqrt{3} \approx 8 \times 1.732 \approx 13.86 \, \text{ms}^{-1}$. This is a reasonable sprint speed for a competitive footballer over a 3-second window. \[ \boxed{8\sqrt{3} \, \text{ms}^{-1}} \]