Since that is quite a high launch velocity, comparable to the escape velocity, you have to take into account the decrease in the acceleration of gravity with distance. You can do that by using the correct formula for the potential energy function V(r), where r is the distance from the center of the Earth. At launch, r1 = re = 6378*10^3 m.
The initial velocity in m/s is
V1 = 3333 m/s.
At the highest altitude, V2 = 0
The potential energy can be written
V(r) = - M g'/r
where g' is the value at the earth's surface (re), 9.80 m/s^2
Solve this equation for the largest distance from earth center, r2:
-Mg'[(1/r) - (1/re)] = (1/2) M V1^2
The M's cancel out. Solve for r and subtract re for the altitude.
A projectile is shot straight up from the earth's surface at a speed of 1.20×10^4 km/h.
How high does it go?
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