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Carnegie Mellon University

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Problem 41 Hard Difficulty

(a) If a person can jump a maximum horizontal distance (by using a $45^{\circ}$ projection angle $)$ of 3.0 $\mathrm{m}$ on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is $g / 6$ and $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ (b) Repeat for Mars, where the acceleration due to gravity is 0.38$g$ .

Answer

A) 18$\mathrm { M }$
B) 7.894$\mathrm { M }$

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Video Transcript

so we can say for party the time of flight is fine by Delta y equaling the y initial t This would be plus 1/2 times the acceleration in the UAE direction Times t squared. Now, of course, we know that Delta Y zero So we can simply say that t the time of flight is gonna be two times V. Why initial divided by G. This makes the range equaling the ex multiplied by tea. Or, we can simply say two times the ex initial the y initial divided by G. Now we can then say that the range on the earth would be two times V x initial V. Why initial divided by the gravity on earth, of course. And then the range on the moon would be equaling two times v x initial times v. Why initial divided by gravity on the moon. And so we can say that so we can say for party the time of flight is fine by Delta y equaling v Y initial t. This would be plus 1/2 times the acceleration in the UAE direction Times t squared. Now, of course, we know that delta y zero, So we can simply say that t the time of flight. It's gonna be two times V. Why initial divided by G Um, this makes the than the range on the moon divided by the range on Earth. It's simply gonna be equaling g sub e divided by geese Obama the gravity on the moon Now gravity on the moon is on Lee equaling 1/6 gravity on earth. And so this is gonna equal six. Which means that for party, the range on the moon is equaling six times the range on the Earth. Given that the gravity on move on the moon is 1/6 as strong range equaling the ex multiplied by tea. Or, we can simply say two times the ex initial the U Y. Initial divided by G. Now we can then say that the range on the earth would be two times V x initial V. Why initial divided by the gravity on earth, of course. And then the range on the moon would be equaling two times V x initial times v why initial divided by gravity on the moon. And so we can say that as the gravity on earth and so this would be six times three meters, 18 meters. So the range on the moon to be 18 meters, four parts A now for part B. Rather, it's the same exact thing for Mars. The range on Mars is gonna be divided by the range on Earth. This would equal the gravity of Earth divided by the gravity on Mars. This is now equaling 0.38 Or rather, we can say that the then the range on the moon, divided by the range on Earth, it's simply gonna be equaling g sub e divided by geese Obama the gravity on the moon Now gravity on the moon is on Lee equaling 1/6 gravity on earth. And so this is gonna equal six. Which means that for party, the range on the moon is equaling six times the range on the Earth. Given that the gravity on move on the moon is 1/6 as strong gravity on Mars is equaling 0.38 times the gravity on earth. And so we can then say that the essentially this would be G sub e divided by 0.38 G's of e, which would mean that the range on Mars is equaling 3.0 meters, divided by point as the gravity on earth. And so this would be six times three meters, 18 meters. So the range on the moon would be 18 meters, four part a now for part B. Rather, it's the same exact thing for Mars. The range on Mars is gonna be divided by the range on Earth. This would equal the gravity of Earth divided by the gravity on Mars. This is now equaling 0.38 Or rather, we can say that the 38 and this is gonna give us 7.9 meters. So this would be the range on Mars if the range on earth was three meters and then the range on the moon is 18 meters. If again, the range on the earth is three meters. That is the end of the solution. Thank you for watching

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