# Falling object Kinematics motion

1. Dec 12, 2013

### negation

1. The problem statement, all variables and given/known data

A falling object travels one-fourth of its distance during the last second of its fall. From what height was it dropped?

2. Relevant equations

none

3. The attempt at a solution

yf - yi = vit + 0.5gt^2
0.25x = 0 + 0.5(-9.8ms^-2)(1)^2
x = 19.6m

2. Dec 12, 2013

### mishek

Are you sure that initial velocity is zero?

3. Dec 12, 2013

### negation

It was not specified so I went in with the assumption vi = 0.

4. Dec 12, 2013

### mishek

The problem states that you have a falling object which has one more second to go (last second of it's fall). That means that from that point initial velocity isn't zero.

5. Dec 12, 2013

### negation

Make sense-should I solve for vf first then?

6. Dec 12, 2013

### mishek

you have two unknowns (T & H), you need two equations.

I would make them for two positions shown on a picture (H and 3/4H) by using above mentioned equation:

H=Vo*t + 1/2*g*t^2

#### Attached Files:

• ###### falling object.png
File size:
2 KB
Views:
126
7. Dec 12, 2013

### negation

H = vit + 0.5gt^2
1/4H = vit + 0.5gt^2

1/4H = vi(1) + 0.5g(1)^2
vi = 1/4H + 4.9ms^-2
H = (1/4H + 4.9ms^-2)t + 0.5gt^2

8. Dec 12, 2013

### mishek

Hello, please check attached photo. Does that make sense?

#### Attached Files:

• ###### falling object.png
File size:
2.8 KB
Views:
146
9. Dec 12, 2013

### negation

Why is vi=0 along the 3/4H?

10. Dec 12, 2013

### mishek

Because object starts falling from height H (from rest) where he has initial velocity 0.