Question

A car moving at 30 m/s makes a head-on collision with a stone wall. From what height would the car have to fall in order to make an equally hard collision with the ground (i.e., hit at the same speed)?

Answer 1

Answer: 45.9 meters

Explanation: Answering this using algebra based physics, we use the equation
`x = x_(0) + v_(y0)*t +(1)/(2) a_(y)t^2`

initial velocity in the y direction is 0, since it is being dropped from rest. the initial position is unknow, as we are solving for it. The ending position is 0, since it will hit the ground, the acceleration is gravity, so -9.81 m/s2. Finally we do not know how many seconds it will take. So to solve we must use a different equation, v = v0 + at

ending velocity is -30 m/s since it is falling down and beginning is 0 since it is dropped from rest.

-30m/s = 0m/s + -9.81 t

simply divide the speed by the acceleration, giving us a time of 3.06seconds. plugging this in the first equation we get

0 = x0 + 0*t +1/2 * -9.81 * .327^2

-x0 = -45.9

meaning the initial height would be

x0 = 45.9m