# College Physics 2.34 – Deceleration of Airmen jumping from airplanes with no parachute

## Solution:

We are given

$x=3\:m$

$v=0\:m/s$

$v_0=54\:m/s$

We are required to solve for the acceleration. We have,

$v^2=v_0^2+2ax$

$\displaystyle a=\frac{v^2-v_0^2}{2x}$

$\displaystyle a=\frac{\left(0\:m/s\right)^2-\left(54\:m/s\right)^2}{2\left(3m\right)}$

$a=-486\:m/s^2$

The negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second.