Final answer:
To find the average velocity of an object dropped from 200 meters during the first 5 seconds, calculate the change in height and divide by time. The object's height after 5 seconds is 77.5 meters, resulting in an average velocity of -24.5 m/s, which is 24.5 m/s when considering magnitude only.
Step-by-step explanation:
The student's question involves calculating the average velocity of an object that is dropped from a height. The function provided, s(t) = -4.9t2 + 200, gives the height of the object after t seconds. To find the average velocity during the first 5 seconds, we would look at the change in position over the change in time. At t = 0 seconds, s(0) = 200 meters (initial height). At t = 5 seconds, s(5) = -4.9(5)2 + 200 = -4.9(25) + 200 = -122.5 + 200 = 77.5 meters (height after 5 seconds).
Now, we calculate the average velocity, vavg, using the formula vavg = (s(t2) - s(t1)) / (t2 - t1). Substituting the known values gives us vavg = (77.5 m - 200 m) / (5 s - 0 s) = (-122.5 m) / 5 s = -24.5 m/s. The negative sign indicates the direction is downwards. Since the question asks for the magnitude, we take the absolute value, which gives us 24.5 m/s average velocity over the first 5 seconds.