**Explanation**

(Option B) **10 m/s:**

We'll analyze this problem assuming a simplified scenario where:

Friction, air resistance, and other energy losses are ignored.

The athlete's mass remains constant during the vault.

We focus on the conversion of kinetic energy (KE) to gravitational potential energy (GPE) at the peak of the vault.

__Conservation of Mechanical Energy:__

We'll use the principle of conservation of mechanical energy, which states that the total mechanical energy of a closed system (no external work done) remains constant. In this case, the closed system is the athlete. Mathematically, this can be expressed as:

KE (initial) + PE (initial) = KE (final) + PE (final)

Since the athlete reaches maximum height at the peak of the vault, their final kinetic energy (KE_final) is negligible (very close to zero). Additionally, the initial potential energy (PE_initial) is zero as the athlete starts from the ground.

Therefore, the equation simplifies to:

KE (initial) = PE (final)

__Calculating the Required Speed:__

__KE (initial)__: This is the kinetic energy the athlete possesses when they plant the pole. We need to find the speed (v) associated with this energy.

__PE (final):__ This is the gravitational potential energy the athlete has at the peak of the vault, which depends on their mass (m), acceleration due to gravity (g), and the height reached (h = 5 m in this case).

Formulas:

KE (initial) = 1/2 * m * v^{2} (where m is mass and v is speed)

PE (final) = m * g * h (where g is acceleration due to gravity and h is height)

Setting Up the Equation:

Since KE (initial) needs to equal PE (final):

1/2 * m * v^{2} = m * g * h

We can divide both sides by the athlete's mass (m) to cancel it out (assuming it doesn't change during the vault):

1/2 * v^{2} = g * h

Solving for Speed (v):

v^{2} = 2 * g * h

v = sqrt(2 * g * h)

Plugging in the Values:

g (acceleration due to gravity) ≈ 9.8 m/s²

h (height) = 5 m

v = sqrt(2 * 9.8 m/s² * 5 m)

v ≈ 10 m/s

(Option A)** 5 m/s: **This speed is too low. The athlete wouldn't have enough kinetic energy to reach a height of 5 meters.

(Option C & D) **15 m/s and. 20 m/s: **While these speeds would provide enough energy to reach 5 meters (or even higher), they're not the most efficient use of the athlete's energy in this simplified scenario.

**S****ubject:** Physics

**Topic:** Work and Energy

**Subtopic:** Conversion of Kinetic Energy to Potential Energy

**Difficulty Level:** Moderate

**Concept** |
**Explanation** |
**Example** |

Kinetic Energy |
The energy possessed by an object due to its motion. It depends on the object's mass and velocity. |
A moving car has kinetic energy, which increases as its speed increases. |

Potential Energy |
The energy possessed by an object due to its position or configuration. It depends on height or distance. |
A book on a shelf has gravitational potential energy due to its height above the ground. |

Conversion of Kinetic Energy to Potential Energy |
When an object's kinetic energy decreases, its potential energy increases, and vice versa, according to the law of conservation of energy. |
In pole vaulting, the athlete's kinetic energy from running is converted to potential energy as they lift themselves against gravity using the pole. |

__Pole Vault__