Final answer:
Using kinematic equations, the velocity of the ball after 7 seconds of free fall is 68.6 m/s, and its velocity upon impact with the ground is approximately 83 m/s, assuming no air resistance.
Step-by-step explanation:
The question pertains to calculations involving a ball dropped from an observation deck 350 m above the ground, and therefore involves the principles of kinematics and free fall under Earth's gravity. To solve for the velocity of the ball after 7 seconds (A), and for the velocity with which it hits the ground (B), we use the equations of motion under constant acceleration.
Given that the acceleration due to gravity (g) is approximately 9.8 m/s2, the initial velocity (u) is 0 m/s since the ball is dropped, and time (t) is 7 seconds for part A:
- v = u + gt => v = 0 + (9.8 m/s2 * 7 s) => v = 68.6 m/s.
This gives us the velocity of the ball after 7 seconds.
For part B, to find out how fast the ball is traveling when it hits the ground, we would need to know the total time the ball takes to reach the ground. However, using the kinematic equation:
- v2 = u2 + 2gh => v = √(2 * 9.8 m/s2 * 350 m) => v ≈ 83 m/s.
Thus, the ball's velocity when it hits the ground is approximately 83 m/s if we ignore air resistance and other factors that could affect the actual motion of the ball.