v =v(o) –gt,
v(o)/2 = v(o) –gt,
t = v(o)/2•g.
h = v(o)•t –g•t^2/2 = v(o)^2/4•g
Acceleration due to gravity on the Moon is
g = G•M/R^2,
where mass and radius of the Moon are
M = 7.33•10^22 kg, R =1.74•10^6 m
G =6.67300•10-11 m^3•kg^-1• s^-2
A projectile is launched vertically from the surface of the Moon with an initial speed of 1500.
At what altitude is the projectile's speed one-half its initial value?
3 answers
I still don't understand how to solve for h
Acceleration due to gravity on the Moom is
g =6.673 •10-11•7.33•10^22/(1.74•10^6)^2 = =1.62 m/s^2
If v(o) = 1500 m/s (units?), then
h = v(o)•t –g•t^2/2 = v(o)^2/4•g =
1500^2/4•1.62 =3.47•10^5 m
g =6.673 •10-11•7.33•10^22/(1.74•10^6)^2 = =1.62 m/s^2
If v(o) = 1500 m/s (units?), then
h = v(o)•t –g•t^2/2 = v(o)^2/4•g =
1500^2/4•1.62 =3.47•10^5 m