Practice Problems In Physics Abhay Kumar Pdf Review

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

A body is projected upwards from the surface of the earth with a velocity of $20$ m/s. If the acceleration due to gravity is $9.8$ m/s$^2$, find the maximum height attained by the body.

Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ practice problems in physics abhay kumar pdf

$0 = (20)^2 - 2(9.8)h$

At maximum height, $v = 0$

(Please provide the actual requirement, I can help you)

Using $v^2 = u^2 - 2gh$, we get

Given $v = 3t^2 - 2t + 1$