Question:

Estimate by what factor a person can jump farther on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.

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Quick Answer:

$R_M = 6R_E$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. Since we have to assume that the landing is at the same level as the launch when you are making this jump on the Earth or the moon, it means that we can use this range formula and this formula by the way only works when you land at the same height as you start with. So it's handy shortcut though in this question and so the range on the Earth is the initial velocity squared times*sin*of 2 times the angle divided by the acceleration due to gravity on the Earth and the range on the moon is the same numerator but this time divided by the acceleration due to gravity on the moon. And so range on the moon divided by range on Earth is the moon formula times the reciprocal of the Earth range, you know, dividing by the range of the Earth is the same as multiplying by its reciprocal and then this part cancels this part giving us that the ratio of the range on the moon to the range on the Earth is

*g E*over

*g M*. And multiply both sides by range on Earth and then we'll get by what factor the range on the moon is multiplied by the range on the Earth. Okay. And so range on the moon is

*g E*over

*g M*times

*R E*and we are told the acceleration due to gravity on the moon is one-sixth that of acceleration due to gravity on the Earth. So we have the

*g E*'s canceling and so we have 1 divided by 1/6 1 divided by 1/6 is the same as 1 times the reciprocal 1 times 6 over 1 so that's 6 and so the range on the moon should be six times the range on the Earth; you can jump six times farther with the same initial launch velocity and angle.