Post a New Question

Physics

posted by on .

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?

  • Physics - ,

    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

  • Physics - ,

    I still don't understand how to solve for h

  • Physics - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question