Practice Problems In Physics Abhay Kumar Pdf [2025]

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

$= 6t - 2$

At maximum height, $v = 0$

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

At $t = 2$ s, $a = 6(2) - 2 = 12 - 2 = 10$ m/s$^2$ practice problems in physics abhay kumar pdf

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

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

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

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