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