at what altitude above the earth's surface is the free fall acceleration equal to three quarters of its value at the surface ?

Answer 1

g wherever = some constant k/r^2
let r = earth radius
let R = radius with 3/4 g
then we want R-r

g surface = k /r^2
g up = k/R^2

g up/g surface = k/R^2/k/r^2 = r^2/R^2 = 3/4
so
R/r = sqrt 4 /sqrt 3
look up r, radius of earth and you can find R, then subtract.