Kinetic energy of object free fall

[bay]

If an object is dropped from height, h, then gravitational energy is E = mgh, but kinetic energy when object hits the ground is E=(1/2)mv^2. Conservation of energy states that E(g)=E(k). Why does kinetic energy not depend on height object was dropped?

 

[ShayanJ]

It does! Its equal to mgh which is a function of h!

 

[bay]

It does! Its equal to mgh which is a function of h!

How does the h cancel out?

 

[ShayanJ]

How does the h cancel out?

It doesn't. $E_k=\frac 1 2 m v^2$ is the definition of kinetic energy of a particle with mass m and speed v regardless of what forces are applied to it. Any particle with mass m and speed v, has the kinetic energy $E_k$. Now if this particle acquired its kinetic energy by falling from a height h under constant gravitational acceleration g, then the numerical value of $mgh$ and $\frac 1 2 m v^2$ will be equal which means the speed of the particle is $v=\sqrt{2gh}$.