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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Hello my purveyors, I have an enigma that is unfortunately perplexing to me...

The archer wants to show his exemplary skill to his friends, so he takes on the challenge of shooting an arrow through the opposing windows of a train that's moving at 120 feet per second. Standing on a platform, the archer shoots. Luckily for him, the arrow reaches the upper right corner of a window(from his vantage point), and the arrow manages to exit through the opposing window's lower left corner(from his vantage point).

Dimensions: The top of the train is 15 ft from the railroad tracks and the tops of each window are 5 ft away from the top of the train. Each window measures 5ft by 24 ft. The (i.e, vertical)perpendicular distance between both opposing windows is 12 ft.

Questions:

If the arrow took 4 seconds to reach the upper right corner of the first window, what were a) the archer's horizontal distance from the train; b) the height of the platform on which he was standing on, if he was standing vertically when he fired the arrow from his shoulder(shoulder height is 5.5 ft). Finally, find c) the time, in seconds, it took for the arrow to hit the ground the moment the arrow left the train(express this in a fraction.)

Neglect all effects of friction, air resistance, drag, and all components of physics that would be problematic. In other words, treat everything as if they were merely points in space, and assume the arrow travelled linearly.

I have perused the problem several times, and it seems to utilize similar triangles and other interesting triangles.( the arrow travels at 65 ft/sec, though my explanation is handwaving.)

I would appreciate a solution from our sagacious members.

-Mathe

The integral of hope is reality.

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Is it perplexing to anyone?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

It is a good question, someone will help you with it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Do you comprehend it?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Not entirely, as you know physics questions leave me going buh, buh buh...

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Apparently our teacher was so sagacious he could solve this in 10 minutes. I'm utterly confused...

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Has anyone found a solution or some workings to this bewildering problem?

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

My work is convoluted and bound to baffle my colleagues..

*Last edited by Mathegocart (2016-12-28 01:20:04)*

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

Perhaps a 3-D model of the enigma would help us appreciably.... I'll get GeoGebra out and model this.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Post #8 looks like art for your freehand drawing class.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

bobbym wrote:

Hi;

Post #8 looks like art for your freehand drawing class.

Haha, I abhor art and it is an extremely convoluted drawing.

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

I have solved it... with a bombardment of 3d kinematics and a slew of heinous, ugly algebra.

ew.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Can you post your solution?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,082

Mathegocart wrote:

The (i.e, vertical)perpendicular distance between both opposing windows is 12 ft.

-Mathe

I like clarity on this point.Is it not **horizontal** distance between opposing windows?And what is the value of 'g' in fps units? I vaguely remember it as 32 ft/sec^2. Is it right? I am used to S.I. units.

*Last edited by thickhead (2016-12-29 00:26:32)*

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,880

thickhead wrote:

Mathegocart wrote:The (i.e, vertical)perpendicular distance between both opposing windows is 12 ft.

-MatheI like clarity on this point.Is it not

horizontaldistance between opposing windows?And what is the value of 'g' in fps units? I vaguely remember it as 32 ft/sec^2. Is it right? I am used to S.I. units.

I see that clarity is necessary for answering these questions, it is horizontal distance, the force of gravity is approximately equal to 31 ft/sec^2.

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,082

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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