Free Fall
IMAT Interactive Study Tool
What is Free Fall?
Free fall is the motion of an object where gravity is the only force acting upon it. For IMAT purposes, this means we always ignore air resistance.
Because there is a constant net force (gravity), all objects in free fall experience a constant downward acceleration, known as the acceleration due to gravity, \(g\).
Acceleration due to gravity: \(g \approx 9.8 \, m/s^2\)
(For many problems, you can approximate \(g \approx 10 \, m/s^2\))
Kinematic Equations for Free Fall
Free fall is just a special case of motion with constant acceleration. We can use the standard kinematic equations, but we replace the horizontal displacement \(s\) with vertical displacement \(y\), and the acceleration \(a\) with \(-g\) (if we define "up" as the positive direction).
\(v = u - gt\)
\(y = ut - \frac{1}{2}gt^2\)
\(v^2 = u^2 - 2gy\)
Key Concepts of Vertical Motion
For an object thrown vertically upwards:
- The velocity is momentarily zero at the maximum height.
- The acceleration is always \(g\) downwards, even at the maximum height.
- The time taken to reach the maximum height is equal to the time taken to fall back to the initial height.
- The speed at any given height is the same on the way up as it is on the way down.
1. True or False: If you drop a heavy stone and a light feather from the same height in a vacuum, they will hit the ground at the same time.
2. True or False: When a ball is at the very top of its trajectory after being thrown upwards, its acceleration is zero.
Example 1: Dropping a Stone
A stone is dropped from rest from the top of a 45 m tall building. How long does it take to hit the ground? (Use \(g = 10 \, m/s^2\)).
Step 1: Identify the knowns and the unknown.
We define "down" as the positive direction for simplicity. Initial velocity (\(u\)) = 0. Displacement (\(y\)) = 45 m. Acceleration (\(a\)) = \(g = 10 \, m/s^2\). We need to find time (\(t\)).
Step 2: Choose the appropriate kinematic equation.
The equation relating \(y, u, g,\) and \(t\) is \(y = ut + \frac{1}{2}gt^2\).
Step 3: Substitute the values and solve for \(t\).
\(45 = (0)(t) + \frac{1}{2}(10)t^2\)
\(45 = 5t^2\)
\(t^2 = \frac{45}{5} = 9\)
\(t = \sqrt{9} = 3 \, s\)
Conclusion: It takes 3 seconds for the stone to hit the ground.
Example 2: Throwing a Ball Upwards
A ball is thrown vertically upwards with an initial speed of 20 m/s. What is the maximum height it reaches? (Use \(g = 10 \, m/s^2\)).
Step 1: Identify the knowns and the unknown.
We define "up" as the positive direction. Initial velocity (\(u\)) = +20 m/s. At the maximum height, the final velocity (\(v\)) = 0. Acceleration (\(a\)) = \(-g = -10 \, m/s^2\). We need to find the displacement (\(y\)).
Step 2: Choose the appropriate kinematic equation.
The equation relating \(v, u, g,\) and \(y\) is \(v^2 = u^2 - 2gy\).
Step 3: Substitute the values and solve for \(y\).
\(0^2 = (20)^2 - 2(10)y\)
\(0 = 400 - 20y\)
\(20y = 400\)
\(y = \frac{400}{20} = 20 \, m\)
Conclusion: The maximum height the ball reaches is 20 meters.
| General Motion | Free Fall (Up is positive) |
|---|---|
| \(v = u + at\) | \(v = u - gt\) |
| \(s = ut + \frac{1}{2}at^2\) | \(y = ut - \frac{1}{2}gt^2\) |
| \(v^2 = u^2 + 2as\) | \(v^2 = u^2 - 2gy\) |