Final answer:
The initial velocity of the arrow shot straight up that returns to the ground in 8.6 seconds is approximately 42.14 m/s, and it reaches a maximum height of about 90.877 m.
Step-by-step explanation:
When an arrow is shot straight up and comes down in 8.6 seconds, we can calculate the initial speed and the maximum height reached using the laws of motion under gravity.
Assuming acceleration due to gravity (g) is 9.8 m/s² and ignoring air resistance, the arrow's upward journey is half of the total time in air, so it takes 4.3 seconds to reach the top of its trajectory.
Using the formula v = u + at, where v is the final velocity (0 m/s at the highest point), u is the initial velocity, and a is the acceleration, we find the initial velocity u by rearranging the formula to u = v - at, thus u = 0 - (-9.8 m/s² * 4.3 s) giving us approximately 42.14 m/s. The maximum height is found using the formula s = ut + 1/2at², giving s = 42.14 m/s * 4.3 s + 1/2 * (-9.8 m/s²) * (4.3 s)² which results in a height of approximately 90.877 m.