in an air bag test a car traveling at 100km/h is remotely driven into a brick wall. suppose an identical car is dropped onto a hard surface. from what height would the car have to be dropped to have the same impact that with the brick wall?
100 km/h = 100000/3600 = 27.78 m/s
Equate kinetic energy to potential energy at equivalent height, h
mgh = (1/2)mv²
h=v²/(2g)
=27.78²/(2*9.8)
=39.4 m
To determine at what height an identical car would need to be dropped to have the same impact as the car traveling at 100 km/h hitting a brick wall, we can make use of the concept of potential energy and kinetic energy.
Let's assume that the height at which the car needs to be dropped is 'h'. We'll also assume that the car's energy is conserved during the impact, meaning the potential energy gained by being lifted to height 'h' will be converted entirely into kinetic energy upon impact.
1. Calculate the kinetic energy of the car traveling at 100 km/h:
- Convert the velocity of 100 km/h to meters per second (m/s):
100 km/h = (100 * 1000) m / (3600 s) ≈ 27.78 m/s
- Use the formula for kinetic energy: KE = 0.5 * mass * velocity^2
We'll assume the mass of the car is 'm'.
KE1 = 0.5 * m * (27.78 m/s)^2
2. Calculate the potential energy of the car dropped from height 'h':
- The formula for potential energy is PE = mass * gravity * height
We'll assume the acceleration due to gravity is 9.8 m/s^2.
PE2 = m * 9.8 m/s^2 * h
3. Equate the kinetic energy (KE1) and the potential energy (PE2):
0.5 * m * (27.78 m/s)^2 = m * 9.8 m/s^2 * h
4. Simplify the equation:
(27.78 m/s)^2 = 2 * 9.8 m/s^2 * h
27.78^2 = 19.6 * h
5. Solve for 'h':
h = (27.78^2) / 19.6 ≈ 39.38 meters
Therefore, in order for the car dropped from height 'h' to have the same impact as the car traveling at 100 km/h hitting a brick wall, it would need to be dropped from a height of approximately 39.38 meters.