JEE Advance - Physics (2022 - Paper 1 Online - No. 8)
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
0.95d
0.95
- OR
Explanation
$d=\frac{u^{2} \sin 2 \theta}{g}$
$$ H=\frac{u^{2} \sin ^{2} \theta}{2 g} $$
So, after entering in the new region, time taken by projectile to reach ground
$$ \begin{aligned} t &=\sqrt{\frac{2 H}{g^{\prime}}} \\\\ &=\sqrt{\frac{2 u^{2} \sin ^{2} \theta \times 0.81}{2 g \times g}} \\\\ &=\frac{0.94 \sin \theta}{g} \end{aligned} $$
So, horizontal displacement done by the projectile in new region is
$$ \begin{aligned} & x=\frac{0.9 u \sin \theta}{g} \times u \cos \theta \end{aligned} $$
$$ \begin{aligned} x &=\frac{0.9 u \sin \theta}{g} \times u \\\\ &=0.9 \frac{u^{2} \sin 2 \theta}{2 g} \end{aligned} $$
So, $d^{\prime}=\frac{d}{2}+x$
$$ =0.95 d $$
So, $n=0.95 d$
$$ H=\frac{u^{2} \sin ^{2} \theta}{2 g} $$
So, after entering in the new region, time taken by projectile to reach ground
$$ \begin{aligned} t &=\sqrt{\frac{2 H}{g^{\prime}}} \\\\ &=\sqrt{\frac{2 u^{2} \sin ^{2} \theta \times 0.81}{2 g \times g}} \\\\ &=\frac{0.94 \sin \theta}{g} \end{aligned} $$
So, horizontal displacement done by the projectile in new region is
$$ \begin{aligned} & x=\frac{0.9 u \sin \theta}{g} \times u \cos \theta \end{aligned} $$
$$ \begin{aligned} x &=\frac{0.9 u \sin \theta}{g} \times u \\\\ &=0.9 \frac{u^{2} \sin 2 \theta}{2 g} \end{aligned} $$
So, $d^{\prime}=\frac{d}{2}+x$
$$ =0.95 d $$
So, $n=0.95 d$
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