A projectile is launched vertically from the surface of the Moon with an initial speed of 1100 m/s

At what altitude is the projectile's speed one-half its initial value?

I can't seem to get the right answer...please help

Mass of the Moon:
M = 7.35x10^22 kg
Radius of the Moon:
R = 1.738x10^6 m

We will label the projectile's mass as variable m.

G is the universal gravitational constant (6.67x10^-11)

Conservation of energy is probably the easiest way to solve this problem. First find the projectile's kinetic energy and gravitational potential energy.

KE = 1/2*m*v²
KE = 1/2*m*(1100 m/s)²
KE = 605,000m J
GPE = -(G*m*M) / R
GPE = -2,821,000m J
Total initial energy:
605,000m J + -2,821,000m J =
-2,216,000m J

When the projectile has one half of its original speed, it will have 1/4 of its original Kinetic Energy
Final Kinetic Energy = (605,000m J) / 4 =
151,250m J

Now we can use the Final KE and the Total initial energy to solve for the final GPE
-2,216,000m J = 151250m J + GPEf
GPEf = -2,367,250m J

Using the GPE equation we can solve for the distance above the moon's surface (d)
-2,367,250m J = -(G*m*M) / (R + d)
After plugging in the values of G, M, and R, and canceling out the m's, you can solve for d
d = 333,000 m

Final Answer:
333,000 m above the Moon's surface.