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
    and the radius= 1.738 e6
    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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  5. Your way I get
    (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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  6. using your figures
    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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