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 5.66 × 102 m/s. What is the maximum height achieved by the projectile?

USE HINT: Energy is conserved, so   Ki+Ui=Kf+Uf.Ki+Ui=Kf+Uf. Use   Ug=Gm1m2/rUg=Gm1m2/r for potential energy and   K=12mv2K=12mv2 for kinetic energy. Initially the projectile is on the surface of the Moon, so   riri is equal to the Moon's radius. At its highest point the projectile is momentarily at rest, so   vf=0.vf=0.Solve for   rfrf in the conservation of energy equation, then subtract the Moon's radius to get height measured from the surface. The Moon's mass and radius can be found in Appendix C.

Explanation / Answer

Applying energy conservation,

PEi + KEi = PEf + KEf

- G M m / R + m v0^2 /2 = - G M m / r + 0

- (6.67 x 10^-11)(7.348 x 10^22)/(1737 x 10^3) + (5.66 x 10^2)^2 / 2 = - (6.67 x 10^-11)(7.348 x 10^22) / r

r= 1841 x 10^3 m

r = R + h

h= 104.5 x 10^3 m Or 104.5 km

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