# Solving Free-Fall Problems

# Falling From Rest

##

Problem 1 - On the Earth:

A bowling ball falls freely (near the surface of the Earth) from
rest. How far does it fall in 4 seconds, and how fast will it be
going at that time?

### Solution 1 - A "Thinking it Through" Approach:

The acceleration of the ball is g - about 10 m/s^{2}. This
means that the velocity of the ball increases by 10 m/s each second,
so after 4 seconds, its velocity will increase by 10 m/s 4 times - to
**40 m/s**.

The instantaneous velocity of the ball increased uniformly from 0
m/s (when it was at rest) to 40 m/s, so its average velocity during
its fall is half-way between 0 m/s and 40 m/s, which is 20 m/s. If
you have an average velocity of 20 m/s for 4 seconds, you move **80
meters**.

### Solution 2 - A Tabular Approach

Since the acceleration = 10 m/s^{2}, the speed of the ball
increases by 10 m/s each second.

The ball's average speed for the first 4 seconds is the average of
0 m/s and 40 m/s, its starting and ending speeds, and distance =
average speed times time.

So the object will have fallen 80 meters, and its speed will be 40
m/s.

### Solution 3 - An Algebraic Approach

Since the ball started from rest:

v = gt = (10 m/s^{2})(4s) = **40 m/s**

So the object will have fallen 80 meters, and its speed will be 40
m/s.

##

Problem 2 - On Another Planet:

A bowling ball falls freely from rest on a planet where g = 5
m/s^{2}. How far does it fall in 4 seconds, and how fast will
it be going at that time?

### Solution 1 - A "Thinking it Through" Approach:

The acceleration of the ball is g for this planet - about 5
m/s^{2}. This means that the velocity of the ball increases
by 5 m/s each second, so after 4 seconds, its velocity will increase
by 5 m/s 4 times - to **20 m/s**.

The instantaneous velocity of the ball increased uniformly from 0
m/s (when it was at rest) to 20 m/s, so its average velocity during
its fall is half-way between 0 m/s and 20 m/s, which is 10 m/s. If
you have an average velocity of 10 m/s for 4 seconds, you move **40
meters**.

### Solution 2 - A Tabular Approach

Since the acceleration = 5 m/s^{2}, the speed of the ball
increases by 5 m/s each second.

The ball's average speed for the first 4 seconds is the average of
0 m/s and 20 m/s, its starting and ending speeds, and distance =
average speed times time.

So the bowling ball will have fallen 40 meters, and its speed will
be 20 m/s.

### Solution 3 - An Algebraic Approach

Since the ball started from rest:

v = gt = (5 m/s^{2})(4s) = **20 m/s**

So the object will have fallen 40 meters, and its speed will be 20
m/s.

last update September 15, 2000 by JL
Stanbrough