Answer:
[tex]v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}[/tex]
Explanation:
The initial mechanical energy of the object, when it is located at height h above the the planet, is just gravitational potential energy:
[tex]E=U=\frac{GMm}{(R+h)}[/tex]
where
G is the gravitational constant
M is the mass of the planet
m is the mass of the object
R is the radius of the planet
h is the altitude of the object
When the object hits the ground, its mechanical energy will sum of potential energy and kinetic energy:
[tex]E=\frac{GMm}{R}+\frac{1}{2}mv^2[/tex]
where
v is the speed of the object at the ground
Since the mechanical energy is conserved, we can write
[tex]\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}[/tex]
and solving for v, we find
[tex]\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}\\v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}[/tex]
