The ball's height at time t is
y = (20.0 m/s) t - 1/2 g t²
where g is the acceleration due to gravity, with magnitude 9.80 m/s².
Also, recall that
v² - u² = 2 a ∆y
where u is the initial velocity, v is the final velocity, a is the acceleration, and ∆y is the change in height. Let Y be the maximum height. At this height, v = 0, so
- (20.0 m/s)² = 2 (-g) Y
==> Y ≈ 20.408 m
Plug this into the first equation and solve for t :
Y = (20.0 m/s) t - 1/2 (9.80 m/s²) t²
==> t ≈ 2.04 s
The ball's velocity at time t is
v = 20.0 m/s - g t
After t = 3.0 s, its velocity will be
v = 20.0 m/s - (9.80 m/s²) (3.0 s)
v = -9.40 m/s
or 9.40 m/s in the downward direction.
