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 $d^{\prime }=nd$. The value of $n$ is ___

Answer 1

Answer: 0.95

$H_{\max }=\frac{v^2{\sin }^2\theta }{2g};\frac{1}{2}{g}_{eff}{t}^2=H_{\max }$
$\Rightarrow t^2=\frac{2H_{\max }}{g_{eff}};t=\sqrt{\frac{v^2{\sin }^2\theta \times 0.81}{g^2}};t=\frac{0.9v\sin \theta }{g}$
$t^2=\frac{2\times v^2{\sin }^2\theta }{2g\left(\frac{g}{0.81}\right)}$
$d^{\prime }=\text{ New range }=\frac{d}{2}+d_1$
$d_1=v\cos {\theta }^{\circ }t$
$=\frac{v^2{\sin }^2\theta \cos \theta \times 0.9}{g};d^{\prime }=\frac{v^2\sin 2\theta }{2g}+\frac{v^2\sin 2\theta \times 0.9}{2g}$
$=\frac{v^2\sin 2\theta }{g}\left(\frac{1.0}{2}\right)=0.95d$
$n=0.95$