The body falls from a height of 100 m what speed it will have at a height of 40 m

The body falls from a height of 100 m what speed it will have at a height of 40 m from the ground using the law of conservation of energy.

Data: h1 (initial height) = 100 m; h2 (end height) = 40 m, the fall is considered free (V0 = 0 m / s).

Constants: g (acceleration due to gravity) = 10 m / s ^ 2.

According to the law of conservation of energy:

En1 + Ek1 = En2 + Ek2, since V0 = 0 m / s, then Ek1 = 0 J.

m * g * h1 = m * g * h2 + m * V ^ 2/2.

g * h1 = g * h2 + V ^ 2/2.

V ^ 2/2 = g * h1 – g * h2.

V ^ 2 = 2g * (h1 – h2).

V = √ (2g * (h1 – h2)).

Let’s perform the calculation:

V = √ (2 * 10 * (100 – 40)) = √1200 = 34.64 m / s.

Answer: At a height of 40 m, the speed of the body is 34.64 m / s.