# If you jump off a 10-meter cliff on the Moon, and the acceleration of gravity is 1.6 m/s^2, then...

## Question:

If you jump off a 10-meter cliff on the Moon, and the acceleration of gravity is 1.6 {eq}m/s^2 {/eq}, then how fast are you going when your feet hit the bottom of the cliff?

## Conservation of Energy:

When energy is said to be conserved, this means that the total energy of a system is kept constant whatever happens to it. For instance, let us imagine that the system's movement speeds up. This means that the kinetic energy of the system increases. However since the total energy is constant, the potential energy decreases.

Given:

• {eq}\displaystyle h = 10\ m {/eq} is the starting height
• {eq}\displaystyle g = 1.6\ m/s^2 {/eq} is the gravitational acceleration on the moon

When you have not yet jumped from the cliff, the total energy you have is due to the gravitational potential energy from being high up from the ground:

{eq}\displaystyle E = mgh {/eq}

Now at the bottom of the cliff, the height h is practically zero. The total energy in this case is entirely kinetic so:

{eq}\displaystyle E = \frac{1}{2} mv^2 {/eq}

Since energy conservation dictates that the total energy be constant, these two energies should be equal:

{eq}\displaystyle \frac{1}{2} mv^2 = mgh {/eq}

Isolating the speed here gives us:

{eq}\displaystyle v = \sqrt{2gh} {/eq}

Now we substitute:

{eq}\displaystyle v = \sqrt{2(1.6\ m/s^2)(10\ m)} {/eq}

We will get:

{eq}\displaystyle \boxed{v = 5.66\ m/s} {/eq} 