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.
At $t = 2$ s, $a = 6(2) - 2 = 12 - 2 = 10$ m/s$^2$ practice problems in physics abhay kumar pdf
Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ A body is projected upwards from the surface
Using $v^2 = u^2 - 2gh$, we get
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. $v = 0$
At maximum height, $v = 0$