# Deceleration of a man falling off a cliff into snow

1. Sep 12, 2014

### salmayoussef

1. The problem statement, all variables and given/known data

A person can just survive a full-body collision (either to the front, back, or side) which results in a deceleration that is up about 100 g's. (One g is 9.8 m/s/s). At greater deceleration fatal brain damage will likely occur. If a 66.4 kg man falls of a cliff of height 37.8 m but manages to land flat on his back in soft snow, undergoing a constant deceleration of this magnitude, how deep would he be buried in the snow?

I feel like this is very simple and I'm just over thinking it...

dy = 37.8 m
g = -980 m/s2

2. Relevant equations

Possibly one of the constant acceleration equations (not sure which).

3. The attempt at a solution

I tried to think about it as if the man was a car driving horizontally then suddenly begins to decelerate, then tried to find his distance after he breaks. No luck. I still can't figure it out... Any advice would be appreciated! :)

2. Sep 12, 2014

### HallsofIvy

Staff Emeritus
Why would you be "not sure"? In the problem it says "(One g is 9.8 m/s/s)"

You know that v= gt where v is the speed after t seconds and d= (g/2)t^2 where d is the distance traveled in t seconds. How long will it take to fall 37.8 m? (I assume that is the distance to the top of the snowbank.) What will his speed be at that point (as he hits the snow bank).

At -100g= -980 m/s^2 (was that what you meant, not "g= -980 m/s2"?) how long will it take for him to stop (his speed to become 0)? And how far will he have gone through the snow in that time?

3. Sep 14, 2014

### salmayoussef

Well, I tried it out using d = 0.5gt2 and found that the time is 2.78 s.

Then using time, I found that the final velocity was 27.2 m/s. (v = gt)

Then for the snow portion, I tried to find time using the same equation and got 0.028 s which I don't think is right. Waaaaay too fast!

And finally, I used d = 0.5(Vf+Vi)t to find that d = 0.38 m. But it isn't the proper answer! Any advice?

4. Sep 14, 2014

### Staff: Mentor

Show your working so we can see where you may be going wrong. Or confirm that you're right.

5. Sep 14, 2014

### CWatters

The equation d = 0.5gt^2 looks like a variant of the SUVAT equations of motion. This problem is mostly about distance and velocity so why use one that involves time? That's not the wrong approach but it might be worth you looking at some of the others.

http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations