• Support PF! Buy your school textbooks, materials and every day products Here!

More help needed

  • Thread starter lunarskull
  • Start date
  • #1
28
0
this problem deals with freely falling bodies

A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventillation box, which she crushed to a depth of 18.0 in. she suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventillator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.

wow im so lost. how do u use the 17th floor anf dent made in the ventillator??? can someone start all 3 parts for me?
 

Answers and Replies

  • #2
Päällikkö
Homework Helper
515
10
a) For uniform acceleration:
[tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
and [tex]v = v_0 + at[/tex]
Or you can use conservation of energy.
b) How is average acceleration defined?
c) see a)
 
  • #3
28
0
still lost...

i plugged in: (18in)=(1728in)(<---converted to inches)+[tex]v_0[/tex]+(1/2)(9.8m/s^2)(t)

what now?
 
  • #4
Päällikkö
Homework Helper
515
10
You missed two t's.
Calculate the speed just before she collides. That means x = 0.
From the fisrt equation, solve for time. Hint: [itex]v_0[/itex] = 0. You know what a is, what is it?
 
  • #5
28
0
a=-9.8m/s correct?
 
  • #6
Päällikkö
Homework Helper
515
10
That is correct.
 
  • #7
28
0
:cry: what do u mean i missed 2 ts?
 
  • #8
28
0
also, when you plug into the equation, both x and [tex]v_0[/tex]=0 correct?
 
  • #9
Integral
Staff Emeritus
Science Advisor
Gold Member
7,198
55
Notice that you are given distances in inches and feet, if you want to use 9.8 m/s^2 as g, you need to convert all distances to meters.
 
  • #10
28
0
ok, i got a right. now how do u do b? i no that the equation is deltav/delta t
 
  • #11
Päällikkö
Homework Helper
515
10
lunarskull said:
:cry: what do u mean i missed 2 ts?
(18in)=(1728in)(<---converted to inches)+[itex]v_0[/itex]+(1/2)(9.8m/s^2)(t)
Instead of [itex]v_0[/itex] you should have [itex]v_0t[/itex] and instead of (t) you should have [itex]t^2[/itex].


b) Well actually, I suppose this is an easier approach:

[tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
[tex]v = v_0 + at[/tex]
Can you see it?
 

Related Threads for: More help needed

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
976
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
828
  • Last Post
Replies
2
Views
1K
Replies
3
Views
1K
Replies
8
Views
7K
Top