A projectile is fired from horizontal ground with speed v and...

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

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

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

$t = \sqrt{\dfrac{2H}{{g}^{'}}}$$= \sqrt{\dfrac{2\left(\dfrac{{v}^{2}\sin^{2}\theta}{2g}\right)}{\dfrac{g}{0 . 81}}}$

$= \dfrac{0 . 94vsin\theta}{g}$

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

$x = \dfrac{0 . 9vsin\theta}{g} \times vcos\theta$

$= 0 . 9\dfrac{{v}^{2}\sin 2\theta}{2g}$

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

$= \dfrac{{v}^{2}\sin 2\theta}{2g} + 0 . 9\dfrac{{v}^{2}\sin 2\theta}{2g}$

$= 0 . 95d$

So, $n = 0 . 95d$