This **question aims** to find the height of the sprinter where the gravitational potential energy is equal to kinetic energy for the world’s fastest human who can reach the speed of 11m/s. The** kinetic energy** of an object is due to its motion. When work is done on an object by applying a net force that transfers energy, the object accelerates, thereby gaining kinetic energy.

**Kinetic energy** is given by the formula:

\[K=\dfrac{1}{2}mv^2\]

The **potential** of the potential object arises from this **position**. **For example**, a **heavy ball in a demolition machine stores energy when it is high**. This stored potential is called **potential energy**. Depending on the position, the **taut bow** can also conserve energy. **Gravity or gravitational force** can be a huge object in relation to something larger because of the force of gravity. The **potential energy** associated with the field of gravity is released (converted into kinetic energy) as objects cross each other.

**Gravitational potential energy** is given by the formula:

\[U=mgh\]

## Expert Answer

**Speed** is given in the question as:

\[v_{human}=v=11\dfrac{m}{s}\]

**Gravitational potential energy** is given as:

\[U=mgh\]

**kinetic energy** is given as:

\[K=\dfrac{1}{2}mv^2\]

$g$ is given as **gravitational acceleration constant** and its value is given as:

\[g=9.8\dfrac{m}{s^2}\]

To increase the **gravitational potential energy** by an amount **equal** to the **kinetic energy** at full speed, the kinetic energy **must be equal** to the gravitational potential energy.

\[K=U\]

\[\dfrac{1}{2}mv^2=mgh\]

\[\dfrac{v^2}{2}=gh\]

\[h=\dfrac{v^2}{2g}\]

**Plug** the values of the gravity $g$ and speed $v$ into the formula to calculate height.

\[h=\dfrac{11^2}{2\times9.8}\]

\[h=6.17m\]

He needs to **climb** $6.17m$ **above the ground**.

## Numerical Result

The **person needs to climb** $6.17m$ above the ground in order to make **kinetic energy equal to gravitational potential energy**.

## Example

The **world’s fastest humans** can reach speeds of about $20\dfrac{m}{s}$. How high does such a sprinter have to climb to **increase the gravitational potential energy by an amount equal to the kinetic energy at full speed**?

**Speed** is given as:

\[v_{human}=v=20\dfrac{m}{s}\]

**Gravitational potential energy** is given as:

\[U=mgh\]

**kinetic energy** is given as:

\[K=\dfrac{1}{2}mv^2\]

“g” is given as **gravitational acceleration constant** and its value is given as:

\[g=9.8\dfrac{m}{s^2}\]

To increase the **gravitational potential energy** by an amount **equal** to the **kinetic energy** at full speed, the kinetic energy **must be equal** to the gravitational potential energy.

\[K=U\]

\[\dfrac{1}{2}mv^2=mgh\]

\[\dfrac{v^2}{2}=gh\]

\[h=\dfrac{v^2}{2g}\]

**Plug** the values of the gravity $g$ and speed $v$ into the formula to calculate height.

\[h=\dfrac{20^2}{2\times9.8}\]

\[h=20.4m\]

He needs to **climb** $20.4m$ **above the ground**.

The **person needs to climb** $20.4m$ above the ground in order to **make kinetic energy equal to gravitational potential energy**.