Answer:
Statement of the given problem,
A ball is thrown vertically upwards at 19.6 m/s. For its complete trip (up and back down to the starting position), what is its average speed?
Let H & T denote the maximum height (in m) of the given ball & the time (in s) required for it to reach there.
Hence from above data we get following kinematic relations,
T = [19.6 m/s - 0 m/s)]/(9.8 m/s^2) [9.8 m/s^2 = gravitational acceleration (assumed)]
or T = 2 (s)
H = (19.6 m/s)*(2 s) - (1/2)*(9.8 m/s^2)*(2 s)^2
or H = 19.6 (m)
Therefore,
assuming vertically upward & downward motions are identical with respect to time & distance covered by the given ball,
the required average speed of the ball during its travel up and back down to the starting position
= (total travel-distance)/(total travel-time)
= 2*(19.6 m)/[2*(2 s)]
= 9.8 m/s [Ans]