A projectile is launched vertically from the surface of the Moon with an initial

Question

A projectile is launched vertically from the surface of the Moon with an initial speed of 1010 m/s. At what altitude is the projectile’s speed one-half it’s initial value?
The answer is 2.7*10^5 m, but I can’t get this number. Please help me with this! A projectile is launched vertically from the surface of the Moon with an initial speed of 1010 m/s. At what altitude is the projectile’s speed one-half it’s initial value?
The answer is 2.7*10^5 m, but I can’t get this number. Please help me with this!
The answer is 2.7*10^5 m, but I can’t get this number. Please help me with this!

Explanation / Answer

According to the given problem,

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*(1010 m/s)²
KE = 510050m J
GPE = -(G*m*M) / R
GPE = -2821000m J
Total initial energy:
510050m J + -2821000m J =
-2310950m J

When the projectile has one half of its original speed, it will have 1/4 of its original Kinetic Energy
Final Kinetic Energy = (510050m J) / 4 =
127512.5m J

Now we can use the Final KE and the Total initial energy to solve for the final GPE
-2310950m J = 127512.5m J + GPEf
GPEf = -2438462.5m J

Using the GPE equation we can solve for the distance above the moon's surface (d)
-2438462.5m 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 = 272467.66 m

Final Answer:
2.7*105 m above the Moon's surface.

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