At maximum height, $v = 0$
Given $v = 3t^2 - 2t + 1$
$0 = (20)^2 - 2(9.8)h$
(Please provide the actual requirement, I can help you) practice problems in physics abhay kumar pdf
Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ At maximum height, $v = 0$ Given $v
$\Rightarrow h = \frac{400}{2 \times 9.8} = 20.41$ m At maximum height
