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

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.

The answer would be A: 6y
Great explanation below.  Keep studying ole Chap