Physics Help?

Question

I've tried this problem but no success, can someone please explain? Thanks

For a movie scene, an 85.0 kg stunt double falls 12.0 m from a building onto a large inflated landing pad. After touching the landing pad surface, it takes her 0.486 s to come to a stop. What is the magnitude of the average net force on her as the landing pad stops her?

Answer 1

This problem is actually several problems in one!

First, consider the stuntwoman's free fall to the top of the landing pad. She falls from rest a distance of 12 m, presumably on earth, so:

Vf^2 - Vo^2=2ad
Vf^2 - 0 = 2 (9.8)(12)
Vf = 15.3 m/s, this is her speed when she makes contact with the pad.

Now, the pad decelerates her:

Vf=at+Vo
0=a(.486)+15.3
a= -31.5 m/s/s (The minus sign, indicates that she is accelerating in the opposite direction from her initial velocity in this part of the problem, which was down. So, she is accelerating up at 31.5 m/s/s.

Now, you've got a Newton's 2nd Law problem. Draw a free body diagram of the woman, I draw a stick person... Then draw arrows representing the forces acting on her, gravity pulls down, the pad pushes up.... That is it...

Newton's 2nd Law says the sum of the forces forward - the sum of the forces backward = ma.

I always call the direction of the acceleration forward, so I can plug in a positive acceleration... So

Fpad - mg = ma
Fpad - 85(9.8) = 85(31.5)
Fpad = 3510.5 N is the upward force exerted on her by the pad.

Answer 2

After a fall of 12m she is traveling at
v = √(2gy) = √(2*9.8*12) = 15.34 m/s since
v = v0 - at she will get to v = 0 with a v0 of 15.34 in .486 s if a is
15.34/.486 = 31.59 m/s² (about 3.22 g)
Since f = ma the force on her will be
f = 85*31.59 = 2685 N.
A bit of a rough ride, but that's why they get paid the big bucks☺

Hope that helps


Doug