Question

34.
For a ground projectile projected at 45°, the magnitude of the average velocity for the entire time
flight is u. Then, the magnitude of the average velocity for the first one - third of time of flight is​

Answer 1

Answer:

Given = angle = 45° time = u

to find = magnitude of avarage velocity

solution =

we can find the avarage velocity, as

avarage velocity = initial velocity - final velocity / 2

so,  

the avarage velocity in 1st one third is time is = (avarge velocity)/ 3  

u/3