Solving Freely Falling Body Problem: 17th Floor & 18.0 in. Ventillation Box

  • Thread starter lunarskull
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In summary, the woman falls from the 17th floor of a building and lands on a metal ventillation box, which she crushes to a depth of 18 in. She suffered only minor injuries, and neglecting air resistance, her speed and average acceleration are calculated.
  • #1
lunarskull
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 I am so lost. how do u use the 17th floor anf dent made in the ventillator? can someone start all 3 parts for me?
 
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  • #2
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
still lost...

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

what now?
 
  • #4
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
a=-9.8m/s correct?
 
  • #6
That is correct.
 
  • #7
:cry: what do u mean i missed 2 ts?
 
  • #8
also, when you plug into the equation, both x and [tex]v_0[/tex]=0 correct?
 
  • #9
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
ok, i got a right. now how do u do b? i no that the equation is deltav/delta t
 
  • #11
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?
 

1. What is the formula for calculating the velocity of a freely falling body?

The formula for calculating the velocity of a freely falling body is v = gt, where v is velocity in meters per second, g is the acceleration due to gravity (9.8 m/s^2), and t is time in seconds.

2. How do you calculate the distance traveled by a freely falling body?

The formula for calculating the distance traveled by a freely falling body is d = 1/2gt^2, where d is distance in meters, g is the acceleration due to gravity (9.8 m/s^2), and t is time in seconds.

3. Can you solve for the time of a freely falling body if the velocity and distance are known?

Yes, the formula for solving for the time of a freely falling body is t = √(2d/g), where t is time in seconds, d is distance in meters, and g is the acceleration due to gravity (9.8 m/s^2).

4. How does air resistance affect the speed of a freely falling body?

Air resistance can decrease the speed of a freely falling body, as it creates a force opposing the motion of the body. This force, known as drag, increases as the speed of the body increases, eventually reaching a point where it balances out the force of gravity. This is known as terminal velocity.

5. What is the difference between a freely falling body problem and a projectile motion problem?

A freely falling body problem deals with the motion of an object that is only affected by the force of gravity, while a projectile motion problem takes into account other forces, such as air resistance and initial velocity, in addition to gravity. A freely falling body problem also assumes that the object starts from rest, while a projectile motion problem does not have this restriction.

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