If no extra acceleration is added to the rocket, then its velocity at time <em>t</em> is
<em>v</em> = 15 m/s - <em>g t</em>
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.
Also, recall that
<em>v</em>² - <em>u</em>² = 2 <em>a </em>∆<em>x</em>
where <em>u</em> is initial speed, <em>v</em> is final speed, <em>a</em> is acceleration, and ∆<em>x</em> is net displacement.
At the rocket's maximum height ∆<em>x</em>, the velocity is 0. So, the maximum height is
0² - (15 m/s)² = 2 (-<em>g</em>) ∆<em>x</em>
∆<em>x</em> = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m
But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.
As mentioned before, this happens when vertical velocity is 0:
0 = 15 m/s - <em>g t</em>
<em>t</em> = (15 m/s) / (9.80 m/s²) ≈ 1.53 s
