# A Person in Free Fall Newton's 2nd Law

1. Oct 15, 2012

### dacabbo

A Person in "Free Fall" Newton's 2nd Law

1. The problem statement, all variables and given/known data
An 80kg aviator is in free fall and acquires a velocity of 60 m/s and then opens his parachute.
After falling an additional 30m, his velocity has been reduced to 20 m/s. What is the average acceleration of the aviator while his fall is being checked? What is the average force being exerted by the parachute?

2. Relevant equations
Force = Mass * acceleration
Average acceleration = change in velocity/change in time

3. The attempt at a solution
Any help would be appreciated, Dont understand how to go about problem

2. Oct 15, 2012

### HallsofIvy

Staff Emeritus
Re: A Person in "Free Fall" Newton's 2nd Law

If a person has intial speed $v_0$ and constant acceleration $a$ has, after time t, speed $at+ v_0$ and will have gone distance $(a/2)t^2+ v_0t$. So we will fall "an additional 30m" when $(a/2)t^2+ 60t= 30$. Solve the quadratic equation $(a/2)t^2+ 60t- 30= 0[itex] for t. Of course, that will depend on a. Put that value of t into [itex]at+ 60= 20$ and solve that equation for a.

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