Question

An object thrown straight up falls back and is caught at the same place it is launched from. Its time of flight is T; its maximum height is H. Neglect air resistance. What is its average velocity for the second half of the trip

Answer 1

The average velocity for the second half of the trip is -2H/T

Step-by-step explanation:

Given;

time of flight as T

maximum height as H

The maximum height of the flight is given by;

`H = (1)/(2)aT^2\\\\ a = (2H)/(T^2)\\\\ But \ average \ velocity \ is \ given \ by , v = aT\\\\v = ((2H)/(T^2) )T\\\\v = (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 \ (1)/(2) \ of \ time \ of \ flight \ (T) = (T)/(2) \\\\The \ average \ velocity = (-H)/(T/2) = (-2H)/(T)`

Therefore, the average velocity for the second half of the trip is -2H/T