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 ${ms}^{- 2}$ ) is (given g = 9.8 ${ms}^{- 2}$ )

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3.5

5.9

16.3

110.8

answer is A.

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Detailed Solution

y = x $\tan θ - \frac{{gx}^{2}}{2 u^{2} \cos^{2} θ}$

For equal trajectories for the same angle of projection

$\frac{g}{u^{2}} = constant \Rightarrow \frac{9.8}{5^{2}} = \frac{g^{'}}{3^{2}}$

$g^{'} = \frac{9.8 \times 9}{25} = 3.528 m / s^{2} = 3.5 m / s^{2}$

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