Final answer:
The initial velocity of the orange tossed by Chef Andy was approximately 7.35 m/s upward, calculated based on the known acceleration due to gravity and the time in the air. The given options do not exactly match the calculated initial velocity.
Step-by-step explanation:
The correct answer to the question of the orange's velocity when it was tossed by Chef Andy is B) The velocity of the orange was 9.8 m/s upward when tossed. This is because the orange was in the air for a total of 0.75 seconds and landed at the same height. From physics, we know that the acceleration due to gravity is 9.8 m/s² downward. The time it takes for the orange to reach the peak of its toss (where the velocity is 0) would be half of the total time in the air, which is 0.75 seconds / 2 = 0.375 seconds. Thus, using the formula for constant acceleration (v = u + at), and knowing that at the peak the velocity (v) is 0, the acceleration (a) is -9.8 m/s² (negative due to the direction being downwards), and the time (t) to reach the peak is 0.375 seconds, we can calculate the initial velocity (u) as follows: 0 = u + (-9.8 m/s²)(0.375 s). Solving for u gives us u = 9.8 m/s² × 0.375 s = 3.675 m/s upward. Therefore, since in these calculations we have divided the time by half to consider the ascent only, the total initial velocity must be double that value, which gives us an initial velocity of approximately 7.35 m/s upward. Note that the precise 9.8 m/s upward indicated as option B would only be the correct initial velocity if the orange were in the air for 1 second since acceleration due to gravity would impose a 9.8 m/s² change in velocity per second. Therefore, the correct answer should consider this and adjust accordingly, which would not match any of the provided options exactly.