Final answer:
The average force applied to the medicine ball from the ground upon impact is 170,000 N in the upward direction, using the Physics concepts of impulse and change in momentum.
Step-by-step explanation:
The subject of this question is Physics, specifically related to the concepts of impulse and force during a collision. An important fact in Physics is that impulse equals the change in momentum of an object, and it can also be calculated as the average force applied during the time of impact times the duration of the collision.
To find the average force exerted on the medicine ball, we need to determine the change in momentum. The medicine ball changes velocity from -50 m/s (just before impact, considering down as negative) to +35 m/s (just after bouncing up). The change in velocity (Δv) is 35 m/s - (-50 m/s) = 85 m/s. The mass of the ball (m) is 10 kg. Thus, the change in momentum (Δp) is mass times the change in velocity: Δp = m × Δv = 10 kg × 85 m/s = 850 kg·m/s.
Given that the impact lasts 0.005 seconds, the average force (F) applied can be calculated using the impulse formula (Impulse = Force × Time), so F = Δp / Time = 850 kg·m/s / 0.005 s = 170,000 N. Since the ball bounces up after the impact, the direction of the average force is upward.