Answer:
[tex] v=\sqrt{(2GM)(\frac{1}{R+h}-\frac{1}{R})}[/tex]
Explanation:
The initial mechanical energy of the object, when it is suspended at a height h above the planet of radius R and mass M, is just gravitational potential energy, so
[tex]E_i = U_i = \frac{GMm}{R+h}[/tex]
When the object reaches the ground, its mechanical energy is now sum of its kinetic energy Kf and its new gravitational potential energy Uf:
[tex]E_f = K_f+U_f = \frac{1}{2}mv^2 + \frac{GMm}{R}[/tex]
Since the mechanical energy is conserved, we can write
Ei = Ef
So we can write
[tex] \frac{GMm}{R+h] =\frac{1}{2}mv^2 + \frac{GMm}{R} [/tex]
from which we can find an expression for v, the speed of the object when it hits the ground:
[tex] v=\sqrt{(2GM)(\frac{1}{R+h}-\frac{1}{R})}[/tex]
