### Present Remotely

Send the link below via email or IM

CopyPresent to your audience

Start remote presentation- Invited audience members
**will follow you**as you navigate and present - People invited to a presentation
**do not need a Prezi account** - This link expires
**10 minutes**after you close the presentation - A maximum of
**30 users**can follow your presentation - Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

### Make your likes visible on Facebook?

Connect your Facebook account to Prezi and let your likes appear on your timeline.

You can change this under Settings & Account at any time.

# MOON BASKETBALL

Description of moon basketball court and ball

by

Tweet## Cameryn Elahee

on 14 December 2012#### Transcript of MOON BASKETBALL

photo credit Nasa / Goddard Space Flight Center / Reto Stöckli Moon Basketball Mission: NASA is designing a state-of-the-art moon colony suitable for men, women, and children. I am one of the engineers drafted into the “recreation” sector. One of the proposed ideas is to have a basket ball court on the moon to keep the astronauts and their families entertained, and to remind them of Earth. NASA has asked me to write a report on the design of the court, the goal is to make the game as close as possible to the way basketball is played on Earth, taking into account the Earth's gravity is 9.81m/s while the moon's gravity is 1.62m/s The Ball: The standard basketball on earth weighs 623.69g, but since I want a ball that weighs the same on the moon as it does on the earth, so to find the mass of the ball we want on the moon. I have to take in account the mass and the gravity because weight= mass*gravity. Ultimately, I want the weight of the ball on the moon to equal same as it does on earth, which means the mass of the ball on the moon is going to be a lot more than the mass of the ball on earth. Hoop Height I will based the height of the hoop by the way the height is determined on earth, which is set as three times the height a normal person can jump to account for the person's height. I got this from calculating how high a player can jump by it they jump average the initial velocity a person has when they jump for a dunk, which is 4.43m/s.

Δy=Viy^2/2g

Δy=4.43m/s^2/(2)(9.81m/s)

Δy=1m

Basketball hoop is 3.1m high, meaning the hoop is three times the height a person is able to jump on earth, which is the same principle I will apply to a person jumping on the moon. Width of the Court The Earth's normal basket ball width is 22m, which came from this model of a player throwing the ball at an initial speed of 12.23m/s at an angle of 45° to another who catches it at the same height. Since that is the way the width is determined on earth, I will apply the same method to moon with the same numbers, but different gravity. To find the width on earth, we are using the constant acceleration model, the classic Δx=(Viycos(ø)(Δt). However, I don't have time interval; to find the time, I will revert to the constant acceleration model, since the ball lands back in it's initial vertical position,I can neglect Δy from our equation because Δy is 0. So the simplified equation comes out to: Δt=(Viysinø)(2)/gravity. After I have the value for change in time, then I will plug the value into the constant acceleration equation and then use the same equation to find the change in distance on the moon by substituting the gravity. The value for x will be the value I use for the width of the court. Citations: http://wiki.answers.com/Q/How_much_does_a_NBA_basketball_weigh

http://www.windows2universe.org/kids_space/moontemp.html

http://wiki.answers.com/Q/How_cold_is_the_moon

http://www.scholastic.com/teachers/article/moon

Crunching the Numbers: I will solve for the mass of the ball on moon by dividing the weight of the ball on the earth by gravity:

mass=(weight)/(gravity);

mass=(623.69g)/(1.62m/s);

mass=384.5g

So, the mass of "the moon ball" is going to be 384.5g Now It's Time to Plug and Chug: Hoop Height (con.) On the moon, person with an initial velocity of 4.43m/s can jump:

Δy=Viy^2/(2)(g)

Δy=(4.43m/s)^2/(2)(1.62m/s)

Δy=6m

I will multiply that by 3.1 to get a hoop height of 18.3m. The Earth's normal basket ball width is 22m, which came from this calculation

Δt=(Viysin45)(2)/gravity

Δt=(12.23m/s)(sin45°)(2)/9.81m/s

Δt=1.76sec.

Δx=(Vi)(Δt)

Δx=(12.23m/s)(1.76sec)

w=12.5m

For the moon, same formula, however, the gravity is swapped with the moon's.

Δt=(Viysin45)(2)/gravity

Δt=(12.23m/s)(sin45°)(2)/1.62m/s

Δt=10.7sec

Δx=(Vi)(Δt)

Δx=(12.23m/s)(10.7sec)

won the width of the moon's basket ball court is

w=130m Length of the Moon ball Court The length of a basketball court is 25m, which would

be the length the Δx, in the model of the width is

being determined by the projectile motion of a ball on earth. To find Δx for the moon, I will find the velocity of the ball on earth, using v=√(2)(g)(Δ x)

to get the initial velocity of the projectile of a ball thrown across the court, the, I will use the initial velocity to plug into the Δt equation, and once I have the change in time, I can solve for change in x through constant acceleration.

have the Δx, so I'm going the use v=√(2)(g)(Δ x) to get the initial velocity of the projectile of a ball thrown across the court, then I

will use the initial velocity to plug into the Δt equation, and once I have the change in time, I can solve for change in x through constant acceleration. Crunching Numbers Vi=√(2)(g)(Δ x)

Vi=√(2)(9.81m/s)(22m)

Vi=20.8m/s Δt=(Vi)(2)/gravity

Δt=(20.8m/s)(2)/1.62m/s

Δt=25.6sec

Δx=(Vi)(Δt)Δx=(20.8m/s)(25.6sec)

so the length of the moon's basketball court is l=534m Do My Answers Add Up? I assume that on the moon the length, width, and height of the hoop will be a lot bigger than it is on earth because the moon has a substantial less amount of gravity. Furthermore, before I started designing the court on the moon, I wanted the game on the moon to be just it is on earth; however, the game on the moon will be played a lot slower because the gravity is smaller. This principle is evident from this calculator of how long it takes for a ball to fall 1.5m on the earth versus the moon:

Δt= √(2Δy)(-g)

Earth:

Δt= √(2*1.5m)/(-9.81m/s)

Δt=.55sec

Moon:

Δt= √(2*1.5m)/(-1.62m/s)Δt=1.4sec Inside Moonball vs. Outside Moonball Nasa wants to know where they should build the moonball court, inside or outside, so I will list the pros and cons and then give them my recommendation. Inside Moonball:

Pros:

~Controlled temperature

~Controlled lighting

~ Easy access to the rest of the colony

Cons:

~Bland Landscape- the colonists may get bored of seeing the inside of the same building making them want to go outside for a change of scenery.

~More Maintence to keep the facility in shape

~Less space- if the court was outside, then the colonist can play outside the boundaries of the court, but if the court is in side, the colonist can run into walls if they run out of space. Outside Moonball:

Pros:

~Colonist are able to spend time on the surface on the moon, which would be an exciting experience for the astronauts.

~Less Maintenance is required because the court is outside.

~Isolation from the rest of the colony-if a colonist wants to get away the hustle and bustle of colony life, they go to an isolated place to relax.

Cons:

~Temperature- on the moon changes drastically because the moon has no atmosphere. The temperature can be hot enough to boil water or as cold as -280° F

~Sound- there is no sound on the moon because there is no air for sound waves to travel through. Imagine playing basketball with no sound, I personally think sound is an underrated but important element of basketball, so we should provide the colonists with it so they enjoy the game more.

~Moon Craters- It would take more effort to find a flat surface on the moon to play moonball than it would to build a flat surface I recommend the Moonball Court be built inside because the environment can be controlled more effectively, and although building and maintaining an inside court may take more effort, providing the colonist with consistent lighting and temperature is worth it. Cameryn Elahee

7th period

Full transcriptΔy=Viy^2/2g

Δy=4.43m/s^2/(2)(9.81m/s)

Δy=1m

Basketball hoop is 3.1m high, meaning the hoop is three times the height a person is able to jump on earth, which is the same principle I will apply to a person jumping on the moon. Width of the Court The Earth's normal basket ball width is 22m, which came from this model of a player throwing the ball at an initial speed of 12.23m/s at an angle of 45° to another who catches it at the same height. Since that is the way the width is determined on earth, I will apply the same method to moon with the same numbers, but different gravity. To find the width on earth, we are using the constant acceleration model, the classic Δx=(Viycos(ø)(Δt). However, I don't have time interval; to find the time, I will revert to the constant acceleration model, since the ball lands back in it's initial vertical position,I can neglect Δy from our equation because Δy is 0. So the simplified equation comes out to: Δt=(Viysinø)(2)/gravity. After I have the value for change in time, then I will plug the value into the constant acceleration equation and then use the same equation to find the change in distance on the moon by substituting the gravity. The value for x will be the value I use for the width of the court. Citations: http://wiki.answers.com/Q/How_much_does_a_NBA_basketball_weigh

http://www.windows2universe.org/kids_space/moontemp.html

http://wiki.answers.com/Q/How_cold_is_the_moon

http://www.scholastic.com/teachers/article/moon

Crunching the Numbers: I will solve for the mass of the ball on moon by dividing the weight of the ball on the earth by gravity:

mass=(weight)/(gravity);

mass=(623.69g)/(1.62m/s);

mass=384.5g

So, the mass of "the moon ball" is going to be 384.5g Now It's Time to Plug and Chug: Hoop Height (con.) On the moon, person with an initial velocity of 4.43m/s can jump:

Δy=Viy^2/(2)(g)

Δy=(4.43m/s)^2/(2)(1.62m/s)

Δy=6m

I will multiply that by 3.1 to get a hoop height of 18.3m. The Earth's normal basket ball width is 22m, which came from this calculation

Δt=(Viysin45)(2)/gravity

Δt=(12.23m/s)(sin45°)(2)/9.81m/s

Δt=1.76sec.

Δx=(Vi)(Δt)

Δx=(12.23m/s)(1.76sec)

w=12.5m

For the moon, same formula, however, the gravity is swapped with the moon's.

Δt=(Viysin45)(2)/gravity

Δt=(12.23m/s)(sin45°)(2)/1.62m/s

Δt=10.7sec

Δx=(Vi)(Δt)

Δx=(12.23m/s)(10.7sec)

won the width of the moon's basket ball court is

w=130m Length of the Moon ball Court The length of a basketball court is 25m, which would

be the length the Δx, in the model of the width is

being determined by the projectile motion of a ball on earth. To find Δx for the moon, I will find the velocity of the ball on earth, using v=√(2)(g)(Δ x)

to get the initial velocity of the projectile of a ball thrown across the court, the, I will use the initial velocity to plug into the Δt equation, and once I have the change in time, I can solve for change in x through constant acceleration.

have the Δx, so I'm going the use v=√(2)(g)(Δ x) to get the initial velocity of the projectile of a ball thrown across the court, then I

will use the initial velocity to plug into the Δt equation, and once I have the change in time, I can solve for change in x through constant acceleration. Crunching Numbers Vi=√(2)(g)(Δ x)

Vi=√(2)(9.81m/s)(22m)

Vi=20.8m/s Δt=(Vi)(2)/gravity

Δt=(20.8m/s)(2)/1.62m/s

Δt=25.6sec

Δx=(Vi)(Δt)Δx=(20.8m/s)(25.6sec)

so the length of the moon's basketball court is l=534m Do My Answers Add Up? I assume that on the moon the length, width, and height of the hoop will be a lot bigger than it is on earth because the moon has a substantial less amount of gravity. Furthermore, before I started designing the court on the moon, I wanted the game on the moon to be just it is on earth; however, the game on the moon will be played a lot slower because the gravity is smaller. This principle is evident from this calculator of how long it takes for a ball to fall 1.5m on the earth versus the moon:

Δt= √(2Δy)(-g)

Earth:

Δt= √(2*1.5m)/(-9.81m/s)

Δt=.55sec

Moon:

Δt= √(2*1.5m)/(-1.62m/s)Δt=1.4sec Inside Moonball vs. Outside Moonball Nasa wants to know where they should build the moonball court, inside or outside, so I will list the pros and cons and then give them my recommendation. Inside Moonball:

Pros:

~Controlled temperature

~Controlled lighting

~ Easy access to the rest of the colony

Cons:

~Bland Landscape- the colonists may get bored of seeing the inside of the same building making them want to go outside for a change of scenery.

~More Maintence to keep the facility in shape

~Less space- if the court was outside, then the colonist can play outside the boundaries of the court, but if the court is in side, the colonist can run into walls if they run out of space. Outside Moonball:

Pros:

~Colonist are able to spend time on the surface on the moon, which would be an exciting experience for the astronauts.

~Less Maintenance is required because the court is outside.

~Isolation from the rest of the colony-if a colonist wants to get away the hustle and bustle of colony life, they go to an isolated place to relax.

Cons:

~Temperature- on the moon changes drastically because the moon has no atmosphere. The temperature can be hot enough to boil water or as cold as -280° F

~Sound- there is no sound on the moon because there is no air for sound waves to travel through. Imagine playing basketball with no sound, I personally think sound is an underrated but important element of basketball, so we should provide the colonists with it so they enjoy the game more.

~Moon Craters- It would take more effort to find a flat surface on the moon to play moonball than it would to build a flat surface I recommend the Moonball Court be built inside because the environment can be controlled more effectively, and although building and maintaining an inside court may take more effort, providing the colonist with consistent lighting and temperature is worth it. Cameryn Elahee

7th period