Question

A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$, then the new range is $d^{\prime}=n d$. The value of $n$ is ___

Answer 1

$H _{\max }=\frac{ v ^2 \sin ^2 \theta}{2 g } ; \frac{1}{2} g _{ eff } t ^2= H _{\max }$
$ \Rightarrow t ^2=\frac{2 H _{\max }}{ g _{ eff }} ; t =\sqrt{\frac{ v ^2 \sin ^2 \theta \times 0.81}{ g ^2}} ; t =\frac{0.9 v \sin \theta}{ g }$
$t ^2 =\frac{2 \times v ^2 \sin ^2 \theta}{2 g \left(\frac{ g }{0.81}\right)} $
$d ^{\prime} =\text { New range }=\frac{ d }{2}+ d _1 $
$d _1 = v \cos \theta^{\circ} t $
$ =\frac{ v ^2 \sin ^2 \theta \cos \theta \times 0.9}{ g } ; d ^{\prime}=\frac{ v ^2 \sin 2 \theta}{2 g }+\frac{ v ^2 \sin 2 \theta \times 0.9}{2 g } $
$=\frac{ v ^2 \sin 2 \theta}{ g }\left(\frac{1.0}{2}\right)=0.95 d $
$ n =0.95$