A projectile is fired from the surface of the earth with a velocity of $10 \mathrm{~m} \mathrm{~s}^{

Question

A projectile is fired from the surface of the earth with a velocity of $10 \mathrm{~m} \mathrm{~s}^{-1}$ and angle $	heta$ with the horizontal. Another projectile fired from another planet with a velocity of $4 \mathrm{~m} \mathrm{~s}^{-1}$ 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${ }^{-2}$) is (Given $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$)

Professor PC Thomas Classes · Answer

(a) : The equation of trajectory is $y=x 	an 	heta-\frac{g x^{2}}{2 u^{2} \cos ^{2} 	heta}$. \ For identical trajectories, $\frac{g}{u^{2}} = 	ext{constant}$. \ $	herefore \frac{9.8}{10^{2}} = \frac{g'}{4^{2}} \Rightarrow g' = \frac{9.8 	imes 16}{100} = 1.568 \mathrm{~m} \mathrm{~s}^{-2}$.