Question

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain height. What height corresponds to a collision at 130 km/h

Answer 1

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

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