# Free Fall: Concept, Equations, Solved Exercises

The free fall is the vertical movement an object undergoes when he is dropped from a certain height near the surface of the Earth. It is one of the simplest and most immediate movements known: in a straight line and with constant acceleration.

All objects that are dropped, or that are thrown vertically up or down, move with the acceleration of 9.8 m / s 2 provided by Earth’s gravity, regardless of their mass.

This fact may be accepted today without problems. However understanding the true nature of free fall took a while. The Greeks had already described and interpreted it in a very basic way by the 4th century BC.

## Free fall motion equations

Once convinced that the acceleration is the same for all bodies released under the action of gravity, it is time to establish the equations necessary to explain this motion.

It is important to emphasize that air resistance is not taken into account in this first movement model. However, the results of this model are very accurate and close to reality.

In everything that follows the particle model will be assumed, that is, the dimensions of the object are not taken into account, assuming that all the mass is concentrated in a single point.

For a uniformly accelerated rectilinear motion in the vertical direction, the y-axis is taken as the reference axis. The positive sense is taken up and the negative down.

### Kinematic magnitudes

Thus, the equations of position, velocity, and acceleration as a function of time are:

#### Acceleration

a = g = -9.8 m / s 2 (-32 ft / s 2 )

#### Position as a function of time: y (t)

y = y o + v o . t + ½ gt 2

Where y o is the initial position of the mobile and v o is the initial velocity. Remember that in the upward vertical throw the initial velocity is necessarily different from 0.

Which can be written as:

y – y o = v o . t + ½ gt 2

Δy = v o . t + ½ gt 2

With Δ y being the displacement effected by the mobile particle. In units of the International System, both the position and the displacement are given in meters (m).

v = v or + g. t

### Speed ​​as a function of displacement

It is possible to deduce an equation that links the displacement with the speed, without time intervening in it. For this, the time of the last equation is cleared:

$t=\frac{v-v_{o}}{g}$

Δy = v o . t + ½ gt 2

The square is developed with the help of the notable product and terms are regrouped.

This equation is useful when you do not have time, but instead you have speeds and displacements, as you will see in the section on worked out examples.

## Examples of free fall

The attentive reader will have noticed the presence of the initial velocity v o . The previous equations are valid for vertical movements under the action of gravity, both when the object falls from a certain height, and if it is thrown vertically up or down.

When the object is dropped, simply set v o = 0 and the equations are simplified as follows.

### Acceleration

a = g = -9.8 m / s 2 (-32 ft / s 2 )

y = y o + ½ gt 2

v = g. t

### Speed ​​as a function of displacement

v 2 = 2g. Dy

Dy will also be negative, since v 2 must be a positive quantity. This will happen whether you take the origin or zero of the coordinate system at the launch point or on the ground.

If the reader prefers, he can take the downward direction as positive. Gravity will continue to act if it is thought to be + 9.8 m / s 2 . But you have to be consistent with the selected sign convention.

### The vertical throw up

Here, of course, the initial velocity cannot be zero. You have to give the object an impulse to rise. According to the initial speed that is provided, the object will rise to a greater or lesser height.

Of course, there will be an instant when the object momentarily stops. Then the maximum height from the launch point will have been reached. Likewise the acceleration is still g downwards. Let’s see what happens in this case.

#### Calculation of the maximum height reached

Choosing i = 0:

Since gravity always points to the ground in the negative direction, the negative sign is canceled.

#### Calculation of the maximum time

A similar procedure is used to find the time it takes for the object to reach the maximum height.

v = v or + g. t

We make v = 0

v o = – g. t max

Flight time is how long the object lasts in the air. If the object returns to the starting point, the rise time is equal to the descent time. Therefore, the flight time is 2. t max.

Is t max twice the total time the object lasts in air? Yes, as long as the object starts from a point and returns to it.

If the launch is made from a certain height above the ground and the object is allowed to proceed towards it, the flight time will no longer be twice the maximum time.

## Solved exercises

In solving the exercises that follow, the following will be considered:

1-The height from where the object is dropped is small compared to the radius of the Earth.

2-Air resistance is negligible.

3-The value of the acceleration of gravity is 9.8 m / s 2

4-When dealing with problems with a single mobile, preferably y o = 0 is chosen at the starting point. This usually makes the calculations easier.

5-Unless otherwise stated, the vertical upward direction is taken as positive.

6-In the combined ascending and descending movements, the equations applied directly offer the correct results, as long as the consistency with the signs is maintained: upward positive, downward negative and gravity -9.8 m / s 2 or -10 m / s 2 if rounding is preferred (for convenience when calculating).

### Exercise 1

A ball is thrown vertically upward with a velocity of 25.0 m / s. Answer the following questions:

a) How high does it rise?

b) How long does it take to reach its highest point?

c) How long does it take for the ball to touch the surface of the earth after it reaches its highest point?

#### Solution

c) In the case of a level launch: t flight = 2. t max = 2 x6 s = 5.1 s

d) When it returns to the starting point, the velocity has the same magnitude as the initial velocity but in the opposite direction, therefore it must be – 25 m / s. It is easily checked by substituting values ​​into the equation for velocity:

### Exercise 2

A small mail bag is released from a helicopter that is descending with a constant speed of 1.50 m / s. After 2.00 s calculate:

a) What is the speed of the suitcase?

b) How far is the suitcase under the helicopter?

c) What are your answers for parts a) and b) if the helicopter is rising with a constant speed of 1.50 m / s?

#### Solution

##### Paragraph a

When leaving the helicopter, the bag carries the initial velocity of the helicopter, therefore v o = -1.50 m / s . With the indicated time, the speed has increased thanks to the acceleration of gravity:

v = v or + g. t = -1.50 – (9.8 x 2) m / s = – 21.1 m / s

##### Section b

Let’s see how much the suitcase has dropped from the starting point in that time:

Bag: Dy = v o . t + ½ gt 2 = -1.50 x 2 + ½ (-9.8) x 2 2 m = -22.6 m

Is selected and or = 0 at the starting point, as indicated at the beginning of the section. The negative sign indicates that the suitcase has descended 22.6 m below the starting point.

Meanwhile the helicopter has descended at a speed of -1.50 m / s, we assume with constant speed, therefore in the indicated time of 2 seconds, the helicopter has traveled:

Helicopter: Δ y = v or .t = -1.50 x 2 m = -3 m.

Therefore after 2 seconds, the suitcase and the helicopter are separated by a distance of:

d = | -22.6 – (-3) | m = 19. 6 m.

Distance is always positive. To highlight this fact, the absolute value is used.

##### Section c

When the helicopter rises, it has a velocity of + 1.5 m / s. With that speed the suitcase comes out, so that after 2 s it already has:

v = v or + g. t = +1.50 – (9.8 x 2) m / s = – 18.1 m / s

The speed turns out to be negative, since after 2 seconds the suitcase is moving downwards. It has increased thanks to gravity, but not as much as in section a.

Now let’s find out how much the bag has descended from the starting point during the first 2 seconds of travel:

Bag: Δ y = v o . t + ½ gt 2 = +1.50 x 2 + ½ (-9.8) x 2 2 m = -16 .6 m

Meanwhile, the helicopter has risen from the starting point, and has done so with constant speed:

Helicopter: Δ y = v or .t = +1.50 x 2 m = +3 m.

After 2 seconds the suitcase and the helicopter are separated by a distance of:

d = | -16.6 – (+3) | m = 19.6 m

The distance that separates them is the same in both cases. The suitcase travels less vertical distance in the second case, because its initial velocity was directed upwards.

## References

1. Kirkpatrick, L. 2007. Physics: A Look at the World. 6 ta Editing abbreviated. Cengage Learning. 23 – 27.
2. Rex, A. 2011. Fundamentals of Physics. Pearson. 33 – 36
3. Sears, Zemansky. 2016. University Physics with Modern Physics. 14 th . Ed. Volume1. 50 – 53.
4. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9 na Ed. Cengage Learning. 43 – 55.
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