To solve this problem, we need to compare the trajectories of two projectiles fired at the same angle but from different planets.$\\$Given:$\\$• Velocity of the projectile on Earth $v_{earth} = 5$ m/s$\\$• Velocity of the projectile on another planet $v_{planet} = 3$ m/s$\\$• Acceleration due to gravity on Earth $g_{earth} = 9.8$ m/s$^2 \\$The range of a projectile is given by the formula:$\\ R = \frac{v^2 \sin(2\theta)}{g} \\$Since the trajectories are identical, the ranges must be equal:$\\ R_{earth} = R_{planet} \\$Substitute the range formula for each projectile:$\\ \frac{v_{earth}^2 \sin(2\theta)}{g_{earth}} = \frac{v_{planet}^2 \sin(2\theta)}{g_{planet}} \\$Cancel $\sin(2\theta)$ from both sides (since $\theta$ is the same for both):$\\ \frac{v_{earth}^2}{g_{earth}} = \frac{v_{planet}^2}{g_{planet}} \\$Rearrange to solve for $g_{planet}: \\ g_{planet} = \frac{v_{planet}^2 \cdot g_{earth}}{v_{earth}^2} \\$Substitute the known values:$\\ g_{planet} = \frac{3^2 \cdot 9.8}{5^2} \\ g_{planet} = \frac{9 \cdot 9.8}{25} \\ g_{planet} = \frac{88.2}{25} \\ g_{planet} = 3.528$ m/s$^2 \\$Rounding to one decimal place, the acceleration due to gravity on the planet is $3.5$ m/s$^2. \\$Therefore, the correct option is Option 1.