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3-2.Motion in Plane
normal
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. . . . .
A$0.40$
B$0.95$
C$0.70$
D$0.80$
(IIT-2022)
Solution
(image)$d=\frac{v^2 \sin 2 \theta}{g}$
(image)
$H _{\operatorname{mex}}=\frac{ v ^2 \sin ^2 \theta}{2 g } ; \frac{1}{2} g _{ aff } t ^2= H _{\max } \Rightarrow t ^2=\frac{2 H _{\max }}{ g _{ af }} ; t =\sqrt{\frac{ v ^2 \sin ^2 \theta \times 0.81}{g^2}} ; t =\frac{0.9 v \sin \theta}{ g }$
$t ^2=\frac{2 \times v ^2 \sin ^2 \theta}{2 g \left(\frac{ g }{0.81}\right)}$
$d ^{\prime}=\text { New range }=\frac{d}{2}+ d _1$
$d _1= v \cos \theta^{\circ} t$
$=\frac{ v ^2 \sin ^2 \theta \cos \theta \times 0.9}{ g } ; d ^{\prime}=\frac{ v ^2 \sin 2 \theta}{2 g }+\frac{ v ^2 \sin 2 \theta \times 0.9}{2 g }$
$=\frac{ v ^2 \sin 2 \theta}{ g }\left(\frac{1.0}{2}\right)=0.95 d$
$n =0.95$
Standard 11
Physics
