A

Step 1: Given the horizontal acceleration of the truck, \( a = 10 \, \text{m/s}^2 \) and the velocity of the truck, \( v_t = 10 \, \text{m/s} \).
Step 2: The ball is thrown with a velocity \( v_b = 10 \, \text{m/s} \) relative to the truck at an angle which we will determine.
Step 3: The geometry of the trajectory suggests this is a projectile motion problem with horizontal and vertical components. Let \( \theta \) be the angle of projection with respect to the horizontal.
Step 4: The motion of the ball follows:

• Horizontal component of the velocity: \( v_{bx} = v_b \cos(\theta) \)
• Vertical component of the velocity: \( v_{by} = v_b \sin(\theta) \)

Step 5: During the time \( t = 1 \, \text{s} \), the horizontal distance traveled as observed by the man on the truck includes the truck's motion and the ball's motion.
The distance by the truck after 1 second is: \( d_t = v_t \cdot t + \frac{1}{2} a imes t^2 = 10 \cdot 1 + \frac{1}{2} \cdot 10 \cdot 1^2 = 10 + 5 = 15 \, \text{m}. \)
Step 6: The horizontal distance covered by the ball: \( d_b = v_{bx} \cdot t = v_b \cos(\theta) \cdot t \).
Step 7: Since both the truck and the ball are moving, we set this distance equal to the total distance observed by the man.
Step 8: Thus, the total distance traveled by the ball in one second is calculated by considering both motion contributions.
Final Calculation: Assuming the ball is thrown at 45 degrees, we find \( d_{total} = d_t + d_b = 15 + (10 \cos(45)) = 15 + 7.07 = 22.07 \, \text{m}.
However, since the components and angle variation were not given, more exploration is needed accordingly. Therefore, in one critical calculation aspect, we assume simplistically that \( v_b (or d_b) aligns as proposed leading to concluding reasons.
Thus, distance of the ball (or effective distance), reasonable options and proximity leads concluding at 15m.
Hence the distance travelled by the ball in one second is to be captured at around 15 m.