LOL !!!
This is a trick question.
You know the answer.
It is exactly the same as throwing a ball straight up.
It takes just as long to go up
(0 = Vi - g t) so t = Vi/g
h = Vi t - (1/2) g t^2
as it takes to go down
0 = h - (1/2) g t^2
= Vi t - (1/2) g t^2 - (1/2) g t^2
t = Vi / g again going down
so t1 = 2 t
a point mass starts moving in a straight line with a constant acceleration a .at a time t from beginning of motion ,the acceleration sign changes, without changing magnitude.determine the time t1from the beginning of motion in which the point of mass returns to initial point
1 answer