A drop of water weighing 0.1 g falls from the roof of a house 36 m high.

A drop of water weighing 0.1 g falls from the roof of a house 36 m high. What would the speed of the drop at the end of the fall be equal to if it fell freely?

Given: m (drop weight) = 0.1 g (0.1 * 10 ^ 3 kg); h (the height of the house from the roof of which the drop was falling) = 36 m; the drop fell freely and V0 (initial drop velocity) = 0 m / s.

Constants: g (acceleration due to gravity) = 9.81 m / s2.

The speed that a drop of water will have at the end of the fall is determined using the law of conservation of mechanical energy (the potential energy of the drop will completely transform into kinetic energy): Ek = En; m * V ^ 2/2 = m * g * h and V = √ (2 * g * h).

Calculation: V = √ (2 * 9.81 * 36) ≈ 26.6 m / s.