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UninstallNow
#1 Feb 7 2010 03:04pm
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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 27 m/s at an angle 25 degree above the horizontal.

How much farther did the ball travel on the moon than it would have on earth?
I can't seem to figure it out even though I tried every way...
ass666
#2 Feb 7 2010 03:13pm
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break the 27m/s into x and y components using sin and cos or whatever is easier for you.
Use the y component and divide it by the acceleration on the moon (9.8/6). That is its up time. Multiply that by two and its its total time for the trajectory until it hits the floor assuming curvature of the moon is negligible.
Get the x component and multiply it by the total air time and there you go =) distance.

Do the same thing but instead of using 9.8/6 use just 9.8 find the distance and take the difference.
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#3 Feb 7 2010 03:30pm
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Quote (ass666 @ Feb 8 2010 06:13am)
break the 27m/s into x and y components using sin and cos or whatever is easier for you.
Use the y component and divide it by the acceleration on the moon (9.8/6). That is its up time. Multiply that by two and its its total time for the trajectory until it hits the floor assuming curvature of the moon is negligible.
Get the x component and multiply it by the total air time and there you go =) distance.

Do the same thing but instead of using 9.8/6 use just 9.8 find the distance and take the difference.


hmm weird this is what i did..
earth
0= vyt+0.5gt^2
gt^2=2vyt
t=2vy/g 2(11.62)/9.8 = 2.37 s

x = (vx+vx)t/2
(24.92+24.92)(2.37)/2 = 59.06m

moon
0= vyt+0.5gt^2
gt^2=2vyt
t=2(vy)/(g/6) 2(11.62)/(9.8/6) = 14.228 s

x = (vx+vx)t/2
(24.92+24.92)(14.22)/2 = 354.4m

so the difference in distance is 295.3 but got it wrong!
what am i doing wrong..? and umm I'm not 100% sure why you have to add 24.92 twice... is it because it's symmtrical?
ass666
#4 Feb 7 2010 03:34pm
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Quote (UninstallNow @ 7 Feb 2010 15:30)
hmm weird this is what i did..
earth
0= vyt+0.5gt^2
gt^2=2vyt
t=2vy/g    2(11.62)/9.8  = 2.37 s

x =  (vx+vx)t/2
(24.92+24.92)(2.37)/2 = 59.06m

moon
0= vyt+0.5gt^2
gt^2=2vyt
t=2(vy)/(g/6)  2(11.62)/(9.8/6) = 14.228 s

x = (vx+vx)t/2
(24.92+24.92)(14.22)/2 = 354.4m

so the difference in distance is 295.3 but got it wrong!
what am i doing wrong..? and umm I'm not 100% sure why you have to add 24.92 twice... is it because it's symmtrical?


I'll check and confirm later, going to super bowl party. I'll do the math there, show you what I did and what I got =)
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#5 Feb 7 2010 03:36pm
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Quote (ass666 @ Feb 8 2010 06:34am)
I'll check and confirm later, going to super bowl party. I'll do the math there, show you what I did and what I got =)


ok thanks.. hopefully I get the response by tomorrow. =)
ass666
#6 Feb 7 2010 04:18pm
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I got the differenceto be 284.9216m

the reson why you added the 24.92 twice is because you divided it by two. If you just left it as was you could have multiplied just Vx*t, you did 2Vs*t/2, the twos cancel out. Yes there is symmetery.

I used lots of floating deicmals in my calculator and thats why I got a different answer than you, but you did it right. Are there any instructions on use so many decimal points, or just to make things easy set the gravity of earth to 10m/s^2? If so follow those instructions.



This post was edited by ass666 on Feb 7 2010 04:22pm
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#7 Feb 7 2010 09:43pm
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Quote (ass666 @ Feb 8 2010 07:18am)
I got the differenceto be 284.9216m

the reson why you added the 24.92 twice is because you divided it by two. If you just left it as was you could have multiplied just Vx*t, you did 2Vs*t/2, the twos cancel out. Yes there is symmetery.

I used lots of floating deicmals in my calculator and thats why I got a different answer than you, but you did it right. Are there any instructions on use so many decimal points, or just to make things easy set the gravity of earth to 10m/s^2? If so follow those instructions.

http://img192.imageshack.us/img192/1143/asdfqe.png


well, it just says to enter answer using 2 sig figs.

hmm.. I put 295 and answer was wrong... so should I just put it as 2.8 * 10^2 make it 280 i guess?

This post was edited by UninstallNow on Feb 7 2010 09:51pm
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#8 Feb 7 2010 10:46pm
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Quote (UninstallNow @ Feb 8 2010 12:43pm)
well, it just says to enter answer using 2 sig figs.

hmm.. I put 295 and answer was wrong... so should I just put it as 2.8 * 10^2 make it 280 i guess?


o god.. i didnt write 25 m/s lol.. no wonder.. it's off by 10

by the way.. I still don't get this part you said "the reson why you added the 24.92 twice is because you divided it by two. If you just left it as was you could have multiplied just Vx*t, you did 2Vs*t/2, the twos cancel out." Is it because the 1 of 24.92 only represents half the the trajectory and other half should be counted as well?

This post was edited by UninstallNow on Feb 7 2010 10:54pm
ass666
#9 Feb 8 2010 12:37pm
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Quote (UninstallNow @ 7 Feb 2010 22:46)
o god.. i didnt write 25 m/s lol.. no wonder.. it's off by 10

by the way.. I still don't get this part you said "the reson why you added the 24.92 twice is because you divided it by two. If you just left it as was you could have multiplied just Vx*t, you did 2Vs*t/2, the twos cancel out." Is it because the 1 of 24.92 only represents half the the trajectory and other half should be counted as well?


you just needed to multiply the velocity along the x axis by the time and that would have been your answer.

You multiplied 2*Vx*t/2

could have just been 24.92*2.37.
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