You can find more problems and solutions like these in the book "Practice Problems in Physics" by Abhay Kumar.
(Please provide the actual requirement, I can help you) practice problems in physics abhay kumar pdf
At maximum height, $v = 0$
$\Rightarrow h = \frac{400}{2 \times 9.8} = 20.41$ m You can find more problems and solutions like
Given $v = 3t^2 - 2t + 1$
Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ practice problems in physics abhay kumar pdf