The friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity. Take g = 10 m/s². The maximum height and time to reach the maximum height will be decreased by

$\begin{array}{l}\mathrm{h}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2{\mathrm{g}}^{\mathrm{\prime }}}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\times \left(\frac{11}{10}\mathrm{g}\right)}\\ =\frac{10}{11}\times \left[\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\right]\\ =\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\times \frac{10}{11}\times 100\mathrm{\%}\\ =91\mathrm{\%}\text{ of }\frac{{\mathrm{u}}^2{\sin }^2\mathrm{e}}{2\mathrm{g}}\end{array}$

So, maximum height decreases by 9%

Further, 

$\begin{array}{l}\mathrm{t}=\frac{\mathrm{usin}\mathrm{\theta }}{{\mathrm{g}}^{\mathrm{\prime }}}=\frac{\mathrm{usin}\mathrm{\theta }}{(11/11)\mathrm{g}}\\ =91\mathrm{\%}\text{ of }\frac{\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}\end{array}$

$\therefore$    time to reach the maximum height decreases by 9%.