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

[astru025]

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.

 

[Simon Bridge]

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.

 

[SteamKing]

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?

 

[lep11]

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.

 

[Nickie]

The correct answer can be found using the equation t = 2v0sin(theta)/g, where t is the time in the air,
v0 is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity on the moon (1.6 m/s^2).
Plugging in the given values, we get t = 2(30m/s)sin(15°)/1.6m/s^2 = 2.5 seconds. Therefore, the ball was in the air for 2.5 seconds before landing back on the surface of the moon.