Answer to Question #137375 in Physics for Godspower

Question

A ball weighing 500n is drop from a height of 30 meters find the kinetic energy of the ball just before hitting the ground.

Answer 1

The ball hits the ground in the time given by the kinematic law:

$$t = \sqrt{\dfrac{2h}{g}}$$

where $h = 30m$ is the height and $g = 9.81m/s^2$ is the gravitational acceleration.

The speed of the ball at this time:

$$v = gt =g\sqrt{\dfrac{2h}{g}}= \sqrt{2gh}$$

The mass of the ball (according to the second Newton's law):

$$m = \dfrac{F}{g}$$

where $F = 500N$ is the weight of the ball.

By definition, the  kinetic energy is:

$$K = \dfrac{mv^2}{2} = \dfrac{F}{g}\cdot \dfrac{\sqrt{2gh}^2}{2} = gFh^2\\ K = 9.81\times 500\times 30^2 = 4.4145 \times 10^6L = 4.4145\space MJ$$

Answer. $4.4145\space MJ.$