# Homework Help: Find deceleration from distance and initial velocity

1. Jul 13, 2013

### -sandro-

1. The problem statement, all variables and given/known data
A man jumps from a building into a safety net located at 11m below, that acts like a rigid body when the man hits it. Find the speed.
The safety net moving 1.5m meters down stops the man completely. Find the deceleration, assumed as constant.

2. Relevant equations

3. The attempt at a solution

This problem is really simple but I'm not sure about the deceleration that seems too high.
I found the speed at 11m (when the man hits the net) of 14.68m/s.
For the other question I set the final velocity at 0m/s, the initial velocity at 14.68m/s, distance 1.5m and I solved for a and I got a (negative) acceleration of 71.34m/s2. Isn't this too high? I tried to find the time with this deceleration to see if I get the given distance but it just won't work.
What do you think?

2. Jul 13, 2013

### Staff: Mentor

That is a reasonable value.
I get a bit more (both with g=9.81m/s^2 and g=10m/s^2), but that could be rounding errors.

If you want to stop a free fall within a fraction of the falling distance, you need the corresponding multiple of the acceleration.

3. Jul 13, 2013

### -sandro-

Hi mfb, are you able to get the distance of 1.5m with that acceleration after finding the time? I can't :|

Edit: nevermind I got it. So apparently that high value is correct!

4. Jul 13, 2013

### Staff: Mentor

I calculated the deceleration directly, without time or velocity.
$$g \frac{11m}{1.5m}=9.81\frac{m}{s^2} \frac{11m}{1.5m} = 71.94\frac{m}{s^2}$$
That equation can be derived via the kinetic energy of the object.

5. Jul 13, 2013

### BruceW

yep, keeping in mind that we are constantly under 9.8m/s^2 all the time, 70m/s^2 is only roughly 7 times greater than that, so it is fairly 'gentle'. In fact, I think astronauts are trained to do like 50m/s^2 for an extended amount of time.

edit: although, as you bounce in the net, you feel as if you are accelerating at 79.8m/s^2 even though you are only at 70 m/s^2 relative to the earth.

p.s. sorry if I just made things more confusing. I was just trying to say that 70m/s^2 is a plausible answer (in the intuitive common-sense way).

Last edited: Jul 13, 2013
6. Jul 14, 2013

Thanks guys!