Example

We consider two crashes:

A car hits a wall at the speed $v$ .
The same car falls from a height $h$ .

How must we choose the height $h$ so that the two crashes are equivalent? We can reason using the concept of energy. In the first case, the kinetic energy of the car is always $E_{k} = \frac{1}{2} m v^{2}$ . In the second one, the kinetic energy of the car when the impact occurs is its initial potential energy $E_{p} = m g h$ . Hence we get : $h = \frac{v^{2}}{2 g}$ . If we take $g = 9.8 m . s^{- 2}$ and $v = 30 m . s^{- 1} = 108 km . h^{- 1} = 67 mph$ , the initial height must be : $h = 46 m = 151 ft$ .

A wall (grey square) at the left bottom corner. A red car at the bottom, going from the left to right where ithit the wall.

E_{k} = \frac{1}{2} m v^{2}

A red car on the right, falling on the wall.

E_{p} = m g h