Explanation

(b) Maximum height, $H = \frac{u^{2}\sin^{2}\theta}{2g}$
$\frac{dH}{du} = \frac{{2u}^{2}\sin^{2}\theta}{2g}\ \ or\ \ dH = \frac{u^{\ }\sin^{2}\theta}{g}du$
$\therefore\ \frac{dH}{H} = \frac{2du}{u}$
Since $H$ is increased by 10%
So, $\frac{dH}{H} = \frac{10}{100} = \frac{1}{10} = \frac{2du}{u}$
Now horizontal range, $dR = \frac{2u}{g}\sin{2\theta}du$
or $\frac{dR}{R} = \frac{2du}{u} = \frac{1}{10}$
$\therefore\ \%$ increase in $R = \frac{dR}{R} \times 100$
$= \frac{1}{10} \times 100 = 10\ \%$