I Love Solving Math Questions

@Romeo Fabulini,

Quote:

Yes, Pilate proclaimed Jesus "Not guilty", but the snooty priests overuled him..

Years before that the Romans leveled Jerusalem.

0 Replies

Okay you maths whizzkids, answer this simple question-
If a high jumper can clear a 7-foot bar on earth, what height would he be able to clear on the moon where the gravity is one-sixth of earths?
It's not exactly a trick question, but you need to know a smidgin about the laws of physics to be able to answer it, I got it from one of Arthur C. Clarkes science books.

1 Reply

@dalehileman,

Yet, what
Nobody at all interested in the math of infinity

0 Replies

@Romeo Fabulini,

How high is the jumper's center of mass?

0 Replies

Quote:

Markr asked: How high is the jumper's center of mass?

You're on the right track but no more clues.. Smile

Come on you maths eggheads, it's a simple enough question-
If a high-jumper can jump 7 feet on earth, how high would he be able to jump on the moon where the gravity is one-sixth less?
You needn't be exact, just an "about" figure will do.
When all your guesses are in (or if nobody can make a guess within the next few days) I'll scan and post the paragraph from Clarke's book that gives the answer.. Smile

2 Replies

@Romeo Fabulini,

Romeo Fabulini wrote:

Quote:
Markr asked: How high is the jumper's center of mass?

You're on the right track but no more clues..
Come on you maths eggheads, it's a simple enough question-
If a high-jumper can jump 7 feet on earth, how high would he be able to jump on the moon where the gravity is one-sixth less?
You needn't be exact, just an "about" figure will do.
When all your guesses are in (or if nobody can make a guess within the next few days) I'll scan and post the paragraph from Clarke's book that gives the answer..

My answer: Higher! No need for the about or approximately...I will stick with "higher."

0 Replies

@Romeo Fabulini,

Assume jumper height is 6 ft. Then assume center of mass of jumper is 3 ft.
So center of mass moves 7 -3 ft = 4 ft on Earth. Assume same energy expended on Earth is expended on Moon. (Let hi and hf be the initial and final height on Earth, h'i and h'f are the heights on the moon, m is the mass of the jumper, and g is the acceleration due to Earth's gravity.)

Ee = Em
mg(hf - hi) = m(1/6)g(h'i - h'f)
mg(4) = (1/6)mg(3-h'f)
24 = (h'f - 3)
27 = h'f

The height jumped on the moon would be 27 feet? That is alot. I doubt it would be that much in a space suit.

1 Reply

@tomr,

Yes you've cracked it at 27 feet which is close enough to Clarkes "around 30 feet", even though the logical answer (trap) would have seemed to be 42 feet!
The key of course is to do with the jumpers centre of gravity.. Wink

Here's the bit from AC Clarkes 'Greetings Carbon-Based Bipeds', page 241-

http://i53.photobucket.com/albums/g64/PoorOldSpike/Photos4-newPB/clarkeA_zpsd274befc.jpg~original

0 Replies

@harpazo,

it figures.

0 Replies

Thank you everyone. I truly appreciate your input.

0 Replies

Thank you everyone.

0 Replies

@harpazo,

How about this type of integral question? (cos^m(x) cos^n(x)) with upper limit is (pi/2), lower limit is 0 dx. I need help on this question, what is the step-by-step working solution.

0 Replies

I Love Solving Math Questions

Related Topics

Quick Links

My Account

able2know