Question

Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m.
Part A
Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit it only 32 m, estimate the acceleration due to gravity on the surface of the Moon. (We neglect air resistance in both cases, but on the Moon there is none.)
Express your answer to two significant figures and include the appropriate units.
gMoon =

Answer 1

@ganeshie8

Answer 2

@CarlosGP

Answer 3

For any acceleration, range (d) can be calculated like this:\[v_y=v_0\sin(\theta)-at \rightarrow t=\frac{ v_0\sin(\theta) }{ a }\]\[d=v_0\cos(\theta)·(2t)=\frac{ v_0^2 \cos(\theta)·2\sin(\theta) }{ a }=\frac{ v_0^2 }{ a }\sin(2 \theta)\]On Earth we can say:\[d_E=\frac{ v_0^2 }{ g}\sin(2 \theta)\]And for the Moon:\[d_M=\frac{ v_0^2 }{ a_M }\sin(2 \theta)\]Therefore:\[\frac{ d_E }{ d_M }=\frac{ a_M }{ g }\rightarrow a_M=\frac{ d_E }{ d_M }g\]

Answer 4

It shows that acceleration on the Moon is something like one sixth the acceleration on Earth, something we already knew

Answer 5

yeah but dont we have to plug in #s for this? so would it be 32/180 X 9.81

Answer 6

Exactly, that is why I said one sixth

Answer 7

i got an answer of 1.744

Answer 8

If you have correctly calculated\[a_M=\frac{ 32 }{ 180 }·9.81\]then it is ok

Answer 9

yes i keep getting 1.744, so it shouldnt be exact #s because we have measurements?

Answer 10

Look\[a_M=\frac{ 32 }{ 180 }·9.81=\frac{ 9.81 }{ 180/32 } =\frac{ 9.81 }{ 5.625 }\approx \frac{ 9.81 }{ 6 }\] The exact calculation is 1.744. That is roughly one sixth (which is not the exact value but it is what we were taught at school and is useful to tell us our result is not preposterous)

Answer 11

ohhh I get it now, but I dont understand how you got the formula for this? is this something I should now bc I would never think of this in the 1st place but i did now the answer was one sixth

Answer 12

Well gravitation on the Moon was well known much before Apollo XI hit the Moon. Therefore no need anybody to get there to play golf. The problem is trying to show how acceleration under the same conditions of speed and angle has an impact on range and range can be used to calculate acceleration likewise.

Answer 13

ohh okay,thank you very much, i had one last question if you are sticking around ?

Answer 14

The imprtant thing is that you understand how the formula of range is calculated

Answer 15

yes, shoot

Answer 16

is there a basic formula for that?

Answer 17

Yes, the formula I have given you:\[d=\frac{ v_0^2\sin(2\theta) }{ a }\]

Answer 18

oh so whenever looking for acceleration or gravity i can use this? does d mean the distance?

Answer 19

exactly, that is the maximum range for a projectile (ball, rocket, cannon ball, whatever) that is launched at a velocity Vo and with an angle=theta

Answer 20

Need to clarify: that formula is valid whenever launch point and hit point are at the same height. If you launch from the top of a hill to impact something that is in valley, it does not work

Answer 21

i didnt understand that..so we dont use it when we are throwing things up or down or falling thinks or jumping things... when do we use it then

Answer 22

You can use whenever the launch point and the landing point of the projectile are at the same height.

Answer 23

can you give me an example please? I still cant picture it in my head

Answer 24

Imagine I am on the street and shoot a cannon. The cannon ball will hit the street at the distance given by the formula.

Now I shoot a cannon from my balcony at the second floor towards the street. That formula is not valid to calculate at what distance (from my balcony) the cannon ball is going to hit the street. My balcony (launch point) and the street (landing or impact point) are not at the same height (level)

Got it now?

Answer 25

ohh okay, so say if i was a ground level and threw something with a lot of force and lands in the same ground level but at a further distance that is when i use this formula for a or g or even v0

Answer 26

exactly, perfect description. Apologies if my explanations are not ok sometimes but English is not my native language

Answer 27

that is alright, it wasnt you, i just need an example sometimes, or all the time lol i am a bit slow with this subject its really difficult for me, and English is not my 1st language either so it makes it more difficult to understand somethings

Answer 28

thank you soo much

Answer 29

u r welcome

Answer 30

i have another question, should i post it here or on other spot, its up to you

Answer 31

I prefer a new spot so everybody can benefit of the thread

Answer 32

okay i will do, thank you