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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 = 9.8 ms⁻².

1. 3.5
2. 5.9
3. 16.3
4. 110.8

Correct answer: 3.5

Solution

The trajectory of a projectile depends on the ratio of the initial velocity squared to the acceleration due to gravity (v²/g). Since the trajectories are identical, the ratio v²/g must be the same for both projectiles. For Earth, (5²/9.8) = (3²/g_planet). Solving for g_planet gives g_planet = 3.5 ms⁻².

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