Problems In Physics Abhay Kumar Pdf: Practice

$= 6t - 2$

Given $u = 20$ m/s, $g = 9.8$ m/s$^2$

(Please provide the actual requirement, I can help you) practice problems in physics abhay kumar pdf

$\Rightarrow h = \frac{400}{2 \times 9.8} = 20.41$ m

Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ $= 6t - 2$ Given $u = 20$ m/s, $g = 9

At $t = 2$ s, $a = 6(2) - 2 = 12 - 2 = 10$ m/s$^2$

At maximum height, $v = 0$

You can find more problems and solutions like these in the book "Practice Problems in Physics" by Abhay Kumar.