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

AI Thread Summary
During Apollo 14, Alan Shepherd hit a golf ball at a 15° angle with an initial velocity of 30 m/s, prompting a calculation of its time in the air. The initial attempt calculated 37.5 seconds based on a misunderstanding of the motion's components and the effect of lunar gravity, which is 1.6 m/s². The discussion emphasized the importance of separating the initial velocity into vertical and horizontal components to accurately determine the time aloft. The incorrect answer highlighted the need to consider the angle of projection in ballistic motion calculations. Ultimately, the correct approach involves using both components of the velocity to find the actual time the ball remained in the air.
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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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