Using conservation of energy, the ball's final velocity is approximately 297.44 m/s when 6.5 x 10^7 m from the Earth's center, considering its initial velocity and gravitational potential energy.
To calculate the speed of the ball when it is 6.5 x 10^7 m from the center of the Earth, we can use the conservation of energy principle, considering both kinetic and gravitational potential energy.
The initial kinetic energy of the ball when thrown by Superman is given by KE_initial = 0.5 * m * v^2, where m is the mass of the ball and v is its initial velocity.
As the ball reaches its highest point, all of its initial kinetic energy is converted into gravitational potential energy (PE_final = mgh), where h is the height above the Earth's surface.
At any point, the total mechanical energy (E) is the sum of kinetic and potential energies: E = KE + PE.
At the highest point, KE_initial = PE_final, so 0.5 * m * v^2 = mgh.
Now, we can use the conservation of energy principle to find the final velocity (v_final) when the ball is 6.5 x 10^7 m from the center of the Earth.
KE_initial = KE_final + PE_final
0.5 * m * v^2 = 0.5 * m * v_final^2 + GMm/r
Where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.
Solving for v_final, we get:
v_final = sqrt(v^2 - 2GM/r)
Substitute the given values: v = 100 m/s, G = 6.67 x 10^-11 N·m^2/kg^2, M = 6.00 x 10^24 kg, r = 6.5 x 10^7 m.
v_final = sqrt(100^2 - 2 * 6.67 x 10^-11 * 6.00 x 10^24 / 6.5 x 10^7)
After calculating, v_final is approximately 297.44 m/s.