How Does Lunar Gravity Affect a Man's Jump Distance Compared to Earth?

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Homework Help Overview

The discussion revolves around the effects of lunar gravity on jump distances, comparing a man's jump on Earth to that on the Moon, where the gravitational acceleration is one-sixth of that on Earth. Additionally, there is a problem involving a basketball player's shot towards a hoop, requiring calculations of initial velocity and projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between gravitational acceleration and jump distance, questioning how these factors influence performance on the Moon compared to Earth. There are attempts to analyze projectile motion in the context of a basketball shot, with considerations of angles and distances involved.

Discussion Status

Some participants have provided feedback on the original poster's attempts, suggesting that there may be inconsistencies in the problem statements. Others have pointed out the importance of understanding the trajectory of the basketball in relation to the hoop and have encouraged further exploration of the physics involved without reaching a consensus on the solutions.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the information they can provide or the methods they can use. There is an ongoing discussion about the assumptions made regarding the jump mechanics and the effects of different gravitational forces.

picklepie159
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Homework Statement



Hi, i have a small problem, can you guys help?
1.
A man jumps a maximum horizontal distance of 3 meters. How far can he jump on the moon, where downward acceleration is g/6?

#2
A 2 meter tall basketball player, 10 meters away from the hoop 3.05 meters high, launches a basketball at a 45 degree angle. What initial velocity must he use to get it in the hoop without touching the rim?


Homework Equations



Vx= vintial*(cos theta)
X-displacement= Vx*T
V-y final= V-initial (sin theta) + AT
X displacement= V * T


The Attempt at a Solution


For number one-
I split it in half- so now I'm trying to solve
A ball is rolled off the edge of a table and lands 1.5 meters away at 45 degrees. What is Vx,
Vy final, T, and height of the table?

So first, I use
Height= 1/2 g t^2
H= 4.9 *t^2

Since D= Vx * T,
Then T= D/vx
So H= 4.9 * D^2/vx^2
Tan 45= Vy/Vx,
Vx= Vy/tan45
D= 1.5 meters
So H= 4.9 * 2.25/ (vy/tan45)
tan45=1
H= 4.9* 2.25/vy^2

and this is where i stop.

For the second question, I was thinking of perhaps the changing it to

A ball is thrown from ground level, at a 45 degree angle, to a hoop 1.05 degrees.
After that, I have no clue

Thanks for answers!

 
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your answer and question for 1st don't match. please check.
 
well started for ques2

now you can understand that the ball must the hoop during second half of its flight.
and also that it will not complete its flight.

so think of it as when the ball reaches point 1.5 along y axis, its x distance should be 10
 
cupid.callin said:
your answer and question for 1st don't match. please check.

well, no I was just looking at it from a different perspective. Instead of a man jumping 3 feet at 45 degrees, I start from his highest point and act as if he is a ball rolling off a table and landing 1.5 meters away at 45 degrees.
 
Good that you know that range is max at 45o.

now using full flight of half flight as you did, you don't need to calculate the max height or final velocities. because they will change when jumping on moon.

Can you tell me what will remain same while jumping on Earth or moon?
 

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