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Q. The friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity. Take $g = 10m/s^2$. The maximum height and time to reach the maximum height will be decreased by

Motion in a Straight Line

Solution:

$H = \frac{u^2}{2g} ; H' = \frac{u^2}{2(g+a)}$
$t_a = \frac{u}{g} ; t^1_a = \frac{u}{(g + a)}$