Question

The kinetic energy needed to project a body of mass m from the earth surface
(radius R) to infinity is
A. ${mgR}/{2}\;$
B. $2mgR$
C. $mgR$
D. ${mgR}/{4}\;$

Answer 1

Hint: The absolute mechanical energy of a ball during projectile motion is its potential energy (PE), and its kinetic energy (KE). (Here we are disregarding any pivot of the ball) without air obstruction, or some other non-moderate force, the absolute energy will be rationed.

Complete answer:
The correct answer is C.
If any object goes through with escaping velocity then the object will never come back at the ground, so we can assume it will reach at infinite distance.
$KE=\dfrac{1}{2}mv_{esc}^{2}=\dfrac{1}{2}m{{(\sqrt{2gR})}^{2}}=mgR$
At the point when a ball is shot straight up, its initial PE is for the most part to be zero since the initial stature is typically characterized to be equivalent to zero, $P{{E}_{i}}=mg{{y}_{i}}$=0. This makes the entirety of the ball's energy related with its motion, for example $K{{E}_{i}}=\dfrac{1}{2}mv_{i}^{2}$. As the ball goes upward the force gravity follows up on the ball quickening it downwards. Since gravity is a traditionalist force it won't change the measure of energy the ball has, yet will make the energy change from Kinetic to potential.

Note: As the ball flies upwards, its velocity will be continually diminishing, again because of the force of the gravity quickening the ball downwards, till inevitably the velocity will be zero, ${{v}_{f}}=0$.
Now the ball will quit moving upwards, in this way the ball will have arrived at its greatest (final) tallness, 𝑦𝑓, and the ball's Kinetic to Potential will be given by, $P{{E}_{f}}=mg{{y}_{f}}$, and $K{{E}_{f}}=\dfrac{1}{2}mv_{f}^{2}=0$.
Since the main force following up on the mass (the ball) during this entire procedure is the force of gravity, the by Conservation of Energy we can set the entirety of the initial energies equivalent to the aggregate of the final energies, giving us; $K{{E}_{i}}+P{{E}_{i}}=K{{E}_{f}}+P{{E}_{f}}$
$\dfrac{1}{2}mv_{i}^{2}+mg{{y}_{i}}=\dfrac{1}{2}mv_{f}^{2}+mg{{y}_{f}}$
$\dfrac{1}{2}mv_{i}^{2}=mg{{y}_{f}}$