A ball is tossed vertically upward with an initial speed of 18.2 m/s. How long does it take before the ball is back on the ground? (Assume the ball is tossed upward from the ground.)

Answer 1

let g = 9.8 m/s^2

v = Vi - g t
0 at top
9.8 t = 18.2
t = 1.86 seconds upward
by symmetry, 1.86 seconds downward
t total = 3.72 s

Answer 2

Oh, you are doing calculus
a = -9.8
then
v = -9.8 t + constant
at t = 0, v = 18.2
so
v = 18.2 - 9.8 t
now height h
h = -(9.8/2)t^2 + 18.2 t + constant
at t = 0, h = 0
h = -4.9 t^2 + 18.2 t
when is h = 0?
t = 0 (of course)
and
t = 18.2/4.9 = 3.71 s close enough :)