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A projectile is fired from the surface of the earth with a velocity of $5 \,m s^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 \,m s^{-1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $\,m s^{-1}$) is
(Given $g = 9.8 \,m s^{-2}$)
$3.5 $
$5.9$
$16.3$
$110.8$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,The\,equation\,of\,trajectory\,is\\
\,\,\,\,\,\,\,\,y = x\tan \theta – \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\\
Where\,\theta \,is\,the\,angle\,of\,projection\,and\\
u\,is\,the\,velocity\,with\,which\,projectile\\
is\,projected.\\
For\,equal\,trajectories\,and\,for\,same\,angles\\
of\,projection,
\end{array}$
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{g}{{{u^2}}} = {\rm{constant}}\\
As\,per\,question,\,\frac{{9.8}}{{{5^2}}} = \frac{{g'}}{{{3^2}}}\\
Where\,g'\,is\,acceleration\,due\,to\,gravity\,on\\
the\,planet.\\
\,\,\,\,\,\,\,\,\,\,\,\,\,g' = \frac{{9.8 \times 9}}{{25}} = 3.5\,m\,{s^{ – 2}}
\end{array}$
