# physics

A projectile is launched vertically from the surface of the Moon with an initial speed of 1210 m/s. At what altitude is the projectile's speed two-fifths its initial value?

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1. Oh well, as a first guess since you did not say I will assume g moon = g earth/6
= 9.8/6 = 1.63 m/s^2
so
a = -1.6
v = Vi - 1.6 t

(2/5)(1210) = 1210 - 1.6 t
1.6 t = 1210 (3/5)
t = 454 seconds
h = Vi t - ((1/2)(1.6) t^2
= 384,447 meters

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2. the mass of the moon is M=7.35e22
I used the formula
1/2m(2/5vi)^2 - GMm/h+R= 1/2vi^2-GMm/R
Where little m cancels out and solved for h. But my answer and the answer you just gave me was marked wrong.
So you have any other suggestions?

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3. well, what did you calculate for g moon? If it is not something like 9.81/6 it is wrong, but the 1/6 of earth gravity is a rough approximation.

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4. 6.67e-11

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(1/2)(21/25) vi^2 = G M [1/R - 1/(R+h) ]

21/50 (1210)^2 = 6.67*10^-11 * 7.35*10^22
[ 1/1.74*10^6 - 1/(1.74*10^6+h) ]

.615*10^6=49*10^11 (.575*10-6 - z)
where
z = 1/(1.74*10^6+h)

.1255*10^-6 = .575*10^-6 - z

z = .449*10^-6
1/z = 2.22*10^6 = 1.74 *10^6 + h
h = .485 *10^6
485,000 meters
check our arithmetics

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g moon = F/m = G M/r^2
= 6.67^10^-11*7.35*10^22 /3.03*10*12
=16.17*10^-1 = 1.67 m/s^2

I guessed 9.8/6 = 1.63 so I would be a little off

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7. okay, I got it thanks !

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8. Great !

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