Practice Problems In Physics Abhay Kumar Pdf New! Today

At maximum height, $v = 0$

$= 6t - 2$

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Acceleration, $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)$ At maximum height, $v = 0$ $= 6t

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

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

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