Answer to Question #185490 in Physics for Kate

Question

15.0kg ball is dropped from 5.00m above ground. What is the speed of the ball just as it hits the ground?

Answer 1

According to the conservation energy law, the kinetic energy of the ball just as it hits the ground is equal to its initial potential energy:

$$E_p = E_k$$

The potential energy can be calculated as follows:

$$E_p = mgh$$

where $m = 13kg$ is the mass of the ball, $g = 9.8m/s^2$ is the gravitational acceleration, and $h = 5m$ is the initial height.

The kinetic energy can be calculated as follows:

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

where $v$ is the required speed of the ball just as it hits the ground. Thus, obtain:

$$mgh = \dfrac{mv^2}{2}\\ v = \sqrt{2gh}\\ v = \sqrt{2\cdot 9.8\cdot 5} \approx 9.9 m/s$$

Answer. 9.9 m/s.