Find deceleration from distance and initial velocity

[-sandro-]

Homework Statement

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.

Homework Equations

V_f2= V_i²+ 2ad

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?

 

[mfb]

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.

 

[-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!

 

[mfb]

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.

 

[BruceW]

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

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).

 

[-sandro-]

Thanks guys!