Question

a ball is dropped from rest at a height of s=71.36 meters above the ground. calculate its speed before it hits the ground

Answer 1

The potential energy of the ball at the height of s=71.36 is the following:

$$E_p = mgs$$

According to the energy conservation law, it will be equal to the kinetic energy of the ball before it hits the ground, which is given by the following expression:

$$E_k = \dfrac{mv^2}{2}$$

Thus:

$$E_k = E_p\\ mgs = \dfrac{mv^2}{2}$$

Expressing the spped, obtain:

$$v^2 = 2gs\\ v = \sqrt{2gs}$$

Substituting the numerical values, get:

$$v = \sqrt{2\cdot 9.81m/s^2 \cdot 71.36m} \approx 18.46m/s$$

Answer. 18.46 m/s.