What is the launch speed of a projectile that rises vertically above the Earth?

i am not too sure

This is solvable by finding the difference between the Projectile's gravitational potential energies at Earth's surface and at a height above the surface of six times Earth's radius, then solving the formula for kinetic energy in terms of velocity for that amount of energy.

Gravitational potential energy is given generally by

U = -GMm/R

U = potential energy, in joules
G = gravitational constant, 6.67 x 10^-11
M = mass of Earth, 5.97 x 10^24kg
m = mass of body at some distance from Earth's center of mass
R = distance of body from Earth's center of mass

Since all we want is a velocity, and gravity accelerates all masses equally, we can define the value of m to be equal to exactly 1, and the equation becomes

U = -GM/R

For U at Earth's surface, we'll let R = r. The difference between the two potential energies is

GM/r - GM/R

r = 6.39 x 10^6m
R = 4.47 x 10^7m

The difference comes out to 5.34 x 10^7J

Kinetic energy is E = mv^2/2. Solving for v this comes to

sqrt[2E/m] = v

Since the projectile mass is defined to be 1kg, this reduces to

sqrt[2E] = v

v = 10,334.41 meters per second

Note: there is a test for the plausibility of this answer. Since the projectile comes to a halt at some finite distance, its initial velocity must be less than the escape velocity of 11,200 meters per second.

Answer 27,386 m/s

How to solve it.

Radius of the Earth = 6,371,000 m. So H = (6,371,000 m) * (6)

H = 38,226,000 m
g = 9.81 m/s^2

Find the time that it takes to rise

t = SQRT { [2H] / g }
t = SQRT { [2 * (38,226,000 m)] / (9.81 m/s^2) }
t = SQRT { [ 76,452,000 m ] / (9.81 m/s^2) }
t = SQRT { 7,793,273 s^2 }
t = 2,792 s

Now find the velocity

Vo = t * g
Vo = (2,792 s) * (9.81 m/s^2)
Vo = 27,386 m/s