Free Fall Physics: Determining Object's Speed Below Balcony Level

When an object is dropped from a certain height, it undergoes free fall under the influence of gravity. In this scenario, the object is dropped from a vertical height of 1.89 m above the balcony level. The question poses a query regarding the object's speed when it descends to a point 2.20 m below the balcony level.

In order to determine the object's speed at the specified point, we can employ the kinematic equation v = √(v0² + 2gh). This equation takes into account the initial speed (v0), acceleration due to gravity (g), and the total vertical distance fallen (h).

The initial speed of the object is 0 m/s since it is dropped from rest. Gravity exerts an acceleration of 9.81 m/s² on the object as it falls. The total distance that the object travels is the sum of the distance fallen before reaching the balcony level (1.89 m) and the additional distance fallen below the balcony level (2.20 m), resulting in a total of 4.09 m.

By substituting these values into the kinematic equation (v0 = 0, g = 9.81 m/s², h = 4.09 m) and solving, we obtain:

v = √(0² + 2 * 9.81 m/s² * 4.09 m) = √(80.0978 m²/s²) ≈ 8.95 m/s

Therefore, the object's speed when it is 2.20 m below the balcony level is approximately 8.95 m/s. This calculation is based on the principles of free fall physics and the kinematic equations governing the motion of objects under gravity.