B

Step 1: Resolve the initial velocity into its components.
The initial speed of the football is given as 20 m/s and the angle of projection is 45º.
The horizontal component of the velocity (Vx) is:
$$V_x = V imes ext{cos}( heta) = 20 imes ext{cos}(45º) = 20 imes \frac{1}{\sqrt{2}} = 10\sqrt{2} ext{ m/s}$$
The vertical component of the velocity (Vy) is:
$$V_y = V imes ext{sin}( heta) = 20 imes ext{sin}(45º) = 20 imes \frac{1}{\sqrt{2}} = 10\sqrt{2} ext{ m/s}$$
Step 2: Calculate the time of flight (T).
The time of flight for projectile motion is given by the formula:
$$T = \frac{2V_y}{g}$$
Substituting the values:
$$T = \frac{2 \times 10\sqrt{2}}{10} = 2\sqrt{2} ext{ seconds}$$
Step 3: Calculate the horizontal distance traveled by the ball.
Using the horizontal component of the velocity, the horizontal distance (D) covered by the ball is:
$$D = V_x \times T = (10\sqrt{2}) \times (2\sqrt{2}) = 20 ext{ meters}$$
Step 4: Calculate the distance the player needs to run.
The player is initially 60 m away, and the ball covers 20 m horizontally. Thus, the distance the player needs to run is:
$$60 - 20 = 40 ext{ meters}$$
Step 5: Calculate the speed of the player (Vp).
The speed of the player can be calculated as:
$$V_p = \frac{ ext{Distance}}{T} = \frac{40}{2\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} ext{ m/s}$$
Therefore, the speed of the second player with which he should run to catch the ball is approximately 14.14 m/s. Hence, the answer is option B.