What was the time in the air for Alan Shepherd's golf ball on the Moon?

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Homework Help Overview

The discussion revolves around the time a golf ball hit by Alan Shepherd during Apollo 14 stays in the air on the Moon. The problem involves projectile motion with an initial velocity of 30 m/s at an angle of 15°, and participants are exploring the implications of lunar gravity on the calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of time in the air, with one suggesting that the total time is derived from the ascent and descent times. Others question the application of the angle of projection and the separation of velocity into components, indicating that the angle affects the trajectory and time aloft.

Discussion Status

The discussion is active, with participants providing insights into the physics of projectile motion. Some have pointed out potential errors in the initial calculations and emphasized the importance of considering the angle of launch. There is no explicit consensus yet, as multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the gravitational acceleration on the Moon is 1.6 m/s², and there is an acknowledgment that the lack of air on the Moon affects the behavior of the golf ball. The initial problem setup and assumptions are under scrutiny, particularly regarding the angle of launch.

astru025
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Homework Statement



During Apollo 14, Alan Shepherd hit a golf ball on the Moon. If he hit the ball at an angle of 15° and an initial velocity of 30m/s,
what was the time in the air?



Homework Equations



30 m/s / 1.6
So it takes 18.750 seconds to get to the top where you start at 0. Then multiply 18.75 by 2 to get 37.5 so you have total time going down and up,

I got 1.6 cause that is gravity on the moon.
37.5 was the incorrect answer though.

The Attempt at a Solution


37.5 did not prove to be correct.
 
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astru025 said:

Homework Statement



During Apollo 14, Alan Shepherd hit a golf ball on the Moon. If he hit the ball at an angle of 15° and an initial velocity of 30m/s,
what was the time in the air?

Homework Equations



30 m/s / 1.6
So it takes 18.750 seconds to get to the top where you start at 0. Then multiply 18.75 by 2 to get 37.5 so you have total time going down and up,
You are arguing that the ball has an initial velocity of 30m/s and it decelerates at 1.6m/s/s along it's direction of travel, changes direction, and then accelerates back to the ground at the same rate. ... however, this cannot be the case. If it did, then the return journey would put it back on the tee ... but it ended up some distance away. But there is a simpler clue: you haven't used the angle part of the initial velocity.

Take another look at your notes for ballistic motion.
When the ball hits the top of it's trajectory, only the vertical component is zero - the horizontal component is constant for the entire motion.
 
astru025 said:

Homework Statement



During Apollo 14, Alan Shepherd hit a golf ball on the Moon. If he hit the ball at an angle of 15° and an initial velocity of 30m/s,
what was the time in the air?



Homework Equations



30 m/s / 1.6
So it takes 18.750 seconds to get to the top where you start at 0. Then multiply 18.75 by 2 to get 37.5 so you have total time going down and up,

I got 1.6 cause that is gravity on the moon.
37.5 was the incorrect answer though.

The Attempt at a Solution


37.5 did not prove to be correct.

Hah! Trick question! There is no air on the moon!

But seriously, what happened to the angle of 15 degrees? Don't you think that makes a difference in how long the golf ball stays aloft?
 
In cases like this when the initial velocity has an angle theta from the horizontal you need to divide the initial velocity into vertical and horizontal components.
 

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