Projectile: questioning need to double time

[Newlander]

Homework Statement

"A ball is thrown straight upward with a velocity of 18 m/s. How much time passes before the ball strikes the ground? Disregard air resistance."

v_i = 18 m/s
a = -9.8 m/s²
t = ?
[not sure: v_f = 0 m/s]

Homework Equations

v_f = v_i + at

The Attempt at a Solution

v_f = v_i + at
t = [(v_f - v_i)/a]
t = [(0 m/s - 18 m/s)/-9.8 m/s²] = 1.8 s

Concerns:
--First, I understand that if I have final velocity as 0 m/s, that refers to when the ball has reached its maximum height--not its pre-impact final velocity.
--Second, with the first comment in mind, I'm not sure my approach is appropriate for this particular problem--and would appreciate guidance if it's not.
--Third and finally, some of my classmates insist that the correct answer is about double the answer I arrived at because "you just multiply the answer by 2 since it's only gone half of the distance" as I have conveyed above. I understand this thinking to some degree, but its simplicity seems to suggest the ball is projected from the ground--rather than simply thrown from what I'm gathering is knee-height. I would love to get your thoughts.

 

[gneill]

There does not appear to be any mention of knees, knee-height, or other initial launch heights in the problem statement, so it's best to assume that it's zero -- launched from ground height.

The ball goes up, decelerating all the way, achieves its maximum height (and zero velocity at that instant), then begins its downward journey, accelerating all the while, until it strikes the ground. By symmetry, it's velocity immediately before striking the ground will be -18m/s.

Your classmates are correct that the point of apogee (highest height achieved) is the halfway point for the entire trajectory.

 

[Newlander]

Thanks very much for the prompt and informative reply!