# Why does a difference in approach to projectile motion yields different results?

A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth, then find the maximum height attained by the body.

If we use Law of Conservation of Energy,

$$\frac{-GMm}{R}+\frac12mu^2=\frac{-GMm}{R+h}+0$$

Here, $\displaystyle u=\sqrt\frac{GM}{2R}$.

On solving this, we get $\displaystyle h=\frac R3$

If we use kinematical equations,

$$v^2=u^2-2gh$$

Here, $v=0,\;\displaystyle u=\sqrt\frac{GM}{2R}=\sqrt\frac{gR}{2}$

$$\therefore \frac{gR}{2}=2gh\implies h=\frac R4$$

Why do we get different answers?

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to Earth's pull.