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? (Use any variable or symbol stated above as necessary.) v2nd half =?

a) 0
b) 2H / T
c) H / T
d) 2H / 2T

Answer 1

The average velocity for the second half of the trip when an object thrown straight up is caught at the same place it is launched from is 0.

Step-by-step explanation:

To find the average velocity for the second half of the trip, we need to understand the motion of the object. When an object is thrown straight up and caught at the same place it is launched from, its time of flight is T and its maximum height is H. Since the object is caught at the same place, the total time of flight is twice the time it takes for the object to reach its maximum height. Therefore, the time for the second half of the trip is T/2. The average velocity for the second half of the trip is equal to the total displacement divided by the time taken. Since the object returns to the same place it was launched from, the total displacement is zero. Therefore, the average velocity for the second half of the trip is 0.