# Free Fall Motion – Equations and Tips

## Summary

Free fall is a uniformly accelerated, rectilinear motion. The object moves vertically and changes under the action of the weight force. On Earth, the acceleration of this type of movement has a modulus of 9.8m/s2

## Equations

The equations for free fall or vertical shot, are those of uniformly accelerated rectilinear movement.

## Introducción

Free fall and vertical throw are basically the same type of movement, a uniformly accelerated rectilinear movement , what changes between them is whether or not they have an initial speed. Let’s remember that this type of movement has a straight path and constant acceleration.

Sometimes, simply, a body is said to have a free fall movement when it is only under the action of the weight force, regardless of whether or not there was initial speed, or whether the object rises or falls.

So if free fall is just a uniformly accelerated rectilinear movement, why complicate things with a bunch of equations? Will it be worth memorizing them all? Will it be worth having to remember in which case each can be applied? It seems more convenient to handle as few equations as possible and to know how to work with them.

Basically any free fall problem can be solved with two equations:

$(1) \hspace{0.5cm} xf = x_{i} + v_{i}.t + \frac{a.t^2}{2}$ $(2) \hspace{0.5cm} a = \frac{\Delta v}{\Delta t}$

From these two equations – the first called “hour law” and the second is the definition of acceleration – we can deduce a third, where time does not appear. This equation, in some cases, allows a more immediate resolution.

$(3) \hspace{0.5cm} v^2_{f} - v^2_{i} = 2.a.\Delta x$

As you can see, they are the equations of rectilinear motion with constant acceleration, and I am not adding any equation to get the maximum height, nor changing the equations according to the direction of motion.

In the case of free fall, the acceleration depends on the planet where the movement is happening. If it is on Earth, the gravitational acceleration has an approximate value of 9.8m/s2

## Exercise: vertical shot from the floor

Let’s start with an easy one. An object is fired from the ground with an initial velocity of 15m / s. Graph position, speed, and acceleration as a function of time for the entire movement, from when it leaves the floor until it returns to the starting position.

Our reference axis “x”, I choose it with the 0 meters located on the floor and the x growing upwards. With this choice, the position of the object (“the x”) matches the height of the object at all times. What’s more, there will be no x negatives, unless the object breaks and crosses the floor!

and, for the moment, nothing else. We could calculate, for example, how long it takes to return to the floor, or calculate the time it takes to climb to the maximum height and double it. I will choose to calculate the total time.

Since I don’t have the final velocity (although it can be easily deduced in this case), I am going to use the hourly law.

$x_{f} = x_{i} + v_{i}.t + \frac{a.t^2}{2}$

The initial position is worth 0 because it starts from the floor, and the final position is also since it returns to the floor.

$0 = 0 + 15m/s.t - \frac{10m/s^2.t^2}{2}$$0 = 15m/s.t - 5m/s^2.t^2$

This is when I need you to like math, easy to solve? The solution time for that equation is 3 seconds. You can verify it.

We need to find out the final speed. Common sense alert: every object you throw with a certain speed will return to your hand at the same speed. Take care of your hands!

So the final velocity is 15m / s, but negative because the object is turning and moving in the opposite direction of the x-axis. But since this is a “training” exercise, let’s try to get to that value from equation (2).

$a = \frac{\Delta v}{\Delta t}$ $\Delta v = a.\Delta t$ $v_{f} - v_{i} = a.\Delta t$ $v_{f} = a.\Delta t + v_{i}$ $v_{f} = -10m/s^2.3,0s + 15m/s$ $v_{f} = -15m/s$

We can make the velocity graph as a function of time.

Also, we already have the necessary information to graph the acceleration.

Finally, to graph the position, we already have a couple of data. For t = 0s it is on the floor (x = 0m), and in 3s the object returned to the floor. As for sketching the parabola, it would be useful for us to calculate the maximum height, which occurs at time t = 1.5s.

$xf = x_{i} + v_{i}.t + \frac{a.t^2}{2}$ $x_{f} = 0 + 15m/s.1,5s - \frac{10m/s^2.1,5^2}{2}$ $x_{f} = 11,25m$

And expressed with the correct number of significant figures is:

$x_{f} = 11m$

Now, we make a graph.

We have finished our first free fall exercise, what a joy! (or not :\)

## Exercises with solutionn

Reference: x = 0m on the floor. Positive direction upwards.

1- A ball is thrown upwards with a speed of 10m / s and from a height of 5.0m. You have to calculate how long it takes to reach the ground and the maximum height the ball reaches.
It takes 2.4 seconds (1.0s to go up and 1.4s to go down), and reaches a maximum height of 10 meters.
2- From a point 55m high, a body is thrown vertically downwards with a speed of 30m / s. Calculate: a) How long does it take to fall? b) How fast do you get to the street?
1,47seconds y -44,7m/s
3- A ball thrown upward returns to the starting point after 6.0s. a) At what speed was it launched? b) How high did it go? c) How far is the ball 4.0s from the ground after it has been thrown?
a) 30m/s b) 45m c) 40m
4- A bricklayer shoots bricks vertically upwards with an initial velocity of 6.0 m / s and from 1.2m from the ground. a) If 0.90 seconds are caught after being thrown, at what height are the bricks caught? b) What speed do they have at that moment? c) How much time do they take to reach the maximum height? d) What is the maximum height reached?