A projectile is fired from horizontal ground with speed v and projection angle θ . When the…

Horizontal range of the projectile is $d=\frac{v^2\sin 2\theta }{g}$

Maximum height attained by the projectile is $H=\frac{v^2{\sin }^2\theta }{2g}$

So, after entering the new region, time taken by projectile to reach ground

$t=\sqrt{\frac{2H}{g^{\text{'}}}}$$=\sqrt{\frac{2\left({\displaystyle \frac{v^2{\sin }^2\theta }{2g}}\right)}{{\displaystyle \frac{g}{0.81}}}}$

$=\frac{0.94v\sin \theta }{g}$

So, horizontal displacement done by the projectile in new region is

$x=\frac{0.9v\sin \theta }{g}\times v\cos \theta$

$=0.9\frac{v^2\sin 2\theta }{2g}$

So, new range is $d^{\text{'}}=\frac{d}{2}+x$

$=\frac{v^2\sin 2\theta }{2g}+0.9\frac{v^2\sin 2\theta }{2g}$

$=0.95 d$

So, $n=0.95 \mathrm{d}$