Explanation

(c) Suppose $t_{0}$ be the time to reach maximum height in the absence of air resistance, then from the relation
$t_{0} = \frac{u\sin\alpha}{g}$ …(i)
When $a$ is retardation caused by air resistance, then total retardation will be $g + a$
$t_{1} = \frac{u\sin\alpha}{g + a} = \frac{u\sin\alpha}{g + \left( \frac{1}{10} \right)g} = \frac{10u\sin\alpha}{11g}$ ...(ii)
Now from equations (i) and (ii), we have
$\therefore t_{1} = \frac{10}{11}t_{0} \Rightarrow t_{0} - t_{1} = t_{0} - \frac{10}{11}t_{0} = \frac{1}{11}t_{0} = 0.09\ t_{0}$
$\therefore\ $Time will decrease by $9\%$