Given $u = 20$ m/s, $g = 9.8$ m/s$^2$
A particle moves along a straight line with a velocity given by $v = 3t^2 - 2t + 1$ m/s, where $t$ is in seconds. Find the acceleration of the particle at $t = 2$ s.
$= 6t - 2$
You can find more problems and solutions like these in the book "Practice Problems in Physics" by Abhay Kumar.
Would you like me to provide more or help with something else? practice problems in physics abhay kumar pdf
$\Rightarrow h = \frac{400}{2 \times 9.8} = 20.41$ m
(Please provide the actual requirement, I can help you) Given $u = 20$ m/s, $g = 9
Given $v = 3t^2 - 2t + 1$
At maximum height, $v = 0$
$0 = (20)^2 - 2(9.8)h$
