Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon's gravity is about 1/6th that of Earth.) A) 6y B) 6/y C) y/6 D) (y + 6) / 2

Question

Rosa

Physics

29 January, 18:19

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon's gravity is about 1/6th that of Earth.) A) 6y B) 6/y C) y/6 D) (y + 6) / 2

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Answers (2)

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Cadence Horton · Answer 1

Since the gravity at the moon is around 6 times weaker, we have that the downwards pull is around 6 times smaller. In equations, we have that g of the moon=g of Earth/6 and since the weight is F=m*g, the force is 1/6th of the force on earth. Hence, the distance is going to be around 6 times larger than the distance traversed on Earth and thus the correct relation is A=6y.

Kayden Ibarra · Answer 2

The answer would be A: 6y

Great explanation below. Keep studying ole Chap