Answer:
The average velocity for the second half of the trip is -2H/T
Explanation:
Given;
time of flight as T
maximum height as H
The maximum height of the flight is given by;
[tex]H = \frac{1}{2}aT^2\\\\ a = \frac{2H}{T^2}\\\\ But \ average \ velocity \ is \ given \ by , v = aT\\\\v = (\frac{2H}{T^2} )T\\\\v = \frac{2H}{T} \ (This \ is \ the \ average \ velocity \ for \ the \ entire \ trip)\\\\During \ the \ second \ half \ of \ the \ motion, the \ height \ of \ travel \ is \ -H \ (negative \ direction)\\\\The \ time \ of \ travel \ is \ \frac{1}{2} \ of \ time \ of \ flight \ (T) = \frac{T}{2} \\\\The \ average \ velocity = \frac{-H}{T/2} = \frac{-2H}{T}[/tex]
Therefore, the average velocity for the second half of the trip is -2H/T
