A projectile is fired from the surface of the earth with a velocity of 5 ms^-1 and angle $θ$ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms^-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 ms^-2 ) (given g = 9.8m/s)²

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3.5

5.9

16.3

110.8

answer is A.

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Detailed Solution

As range = $\frac{u^{2} \sin^{2} θ}{g}$ , So, $g \propto u^{2}$
Therefore, $g_{planet} = {(\frac{3}{5})}^{2} (9.8 m / s^{2}) = 3.5 m / s^{2}$

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