posted by Rachel .
A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface of Earth will it go? (Use any variable or symbol stated above along with the following as necessary: ME for the mass of the Earth, rE for the radius of the Earth, and G for the gravitational constant.)
The equations of accelerated motion, derived elsewhere, also apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g” for the acceleration due to gravity). This results in
....Vf = Vo + gt (the term “g” for acceleration due to gravity is assumed constant)
....d = Vo(t) + g(t^2)
....Vf^2 = Vo^2 + 2gd
where Vo = the initial velocity, Vf = the final velocity, g = the acceleration due to gravity, t = the time elapsed and d = the distance covered during the time period t.
g derives from g = GM/r^2 where
G = the universal gravitational constant,1.069304x10^-9ft.^3/lb.sec.^2,
M = the mass of the earth, 1.31672x10^25 lb., and r = the radius of the earth in feet, 5280(3963.4) = 20,926,752 feet, resulting in a mean g = g = 32.519ft./sec.6@.
As written, these expressions apply to falling bodies.
The equations that apply to rising bodies are
....Vf = Vo - gt (the term “g” for acceleration due to gravity is assumed constant on, or near, the surface of the Earth)
....h = Vo(t)-g(t^2)/2
....Vf^2 = Vo^2-2gh
All of the above ignores surface friction.
The height reached then becomes
h = Vo(t)-32.519(t^2)/2 where
t = (Vo - Vf)32.519 or
h = (Vo^2 - Vf^2)65.04