Previous Year NEET Question- Optics L-7

Question: A projectile is fired from the surface of the earth with a velocity of $5 \mathrm{~m} \mathrm{~s}^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 \mathrm{~m} \mathrm{~s}^{-}$ ${ }^1$ 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 $\mathrm{m} \mathrm{s}^{-2}$ ) is#

(Given $\mathrm{g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}$ )

A) 3.5

B) 5.9

C) 16.3

D) 110.8

Answer: 3.5#

Sol:#

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, $$ \frac{g}{u^2}=\text { constant } $$

According to the question, $\frac{9.8}{5^2}=\frac{g^{\prime}}{3^2}$ where $g^{\prime}$ is acceleration due to gravity on the planet. $$ g^{\prime}=\frac{9.8 \times 9}{25}=3.5 \mathrm{~ms}^{-2} $$