# Falling object Kinematics motion

negation

## Homework Statement

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

none

## The Attempt at a Solution

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

## Answers and Replies

mishek
Are you sure that initial velocity is zero?

negation
Are you sure that initial velocity is zero?

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

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.

negation
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.

Make sense-should I solve for vf first then?

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

#### Attachments

• falling object.png
2 KB · Views: 499
negation
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

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

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

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

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