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 maximu

Question

Travka [436]

4 years ago

13

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 maximu

m height is H. Neglect air resistance. What is its average velocity for the second half of the trip

Physics

1 answer:

Natasha_Volkova [10] · Answer 1 · 2022-02-07T11:32:17+03:00

Natasha_Volkova [10]4 years ago

7 0

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;

$H = \frac{1}{2}aT^2\\\\ a = \frac{2H}{T^2}\\\\ But \ average \ velocity \ is \ given \ by , v = aT\\\\v = (\frac{2H}{T^2} )T\\\\v = \frac{2H}{T} \ (This \ is \ the \ average \ velocity \ for \ the \ entire \ trip)\\\\During \ the \ second \ half \ of \ the \ motion, the \ height \ of \ travel \ is \ -H \ (negative \ direction)\\\\The \ time \ of \ travel \ is \ \frac{1}{2} \ of \ time \ of \ flight \ (T) = \frac{T}{2} \\\\The \ average \ velocity = \frac{-H}{T/2} = \frac{-2H}{T}$

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