Horizontal launch is an example of composition of motion in two dimensions: an u.r.m. on the horizontal axis and a u.a.r.m. in the vertical one. In this section, we will see:

## Concept and representation

Horizontal launch consists of horizontally launching a body , also know as projectile, from a certain height. In the figure below you can see a representation of the situation:

Horizontal launch

Think about a drop that slides at constant velocity (v0) on a leaf located at a height H, when it reaches the edge and falls to the ground. During the fall, it moves at constant velocity v0 in the x-axis (u.r.m.) and it moves in free fall along the y-axis (u.a.r.m.) due to the action of the gravity. Initially, the velocity in this y-axis is 0 (vy = 0) and increases as the projectile descends.
Notice the projections of the motion in the axes and verify that they coincide with the motions that we have described (u.r.m. and u.a.r.m.)

Horizontal launch is the composition of a uniform rectilinear motion (horizontal urm) and a uniformly accelerated rectilinear motion of free fall (vertical uarm). The moving object in this kind of motion is often referred to as projectile, or horizontally launched projectile.

## Equations

The equations for horizontal launch are:

• The equations for the u.r.m. for the x-axis

$x={x}_{0}+{v}_{x}·t$

• The equations of u.a.r.m. for the y-axis

${v}_{y}={v}_{0y}+{a}_{y}·t$

$y={y}_{0}+{v}_{0y}·t+\frac{1}{2}·{a}_{y}·{t}^{2}$

Since, as stated above, the velocity forms an angle α with the horizontal, the components x and y are determined by using the most common trigonometric relationships:

Decomposition of the velocity vector

Any vector, including the velocity, can be broken down is in 2 vectors, vx and vy, that have the same directions as the Cartesian axes. The magnitude of both vectors can be calculated from the angle that the vector forms with the horizontal through the expressions shown in the figure.

Finally, taking into consideration what we previously stated, that y0 = H , x0 = 0 and ay = -g, we can rewrite the formulas as they are shown in the following table. These are the final expressions for the calculation of a horizontally launched projectile kinematic magnitudes:

Position (m) Velocity (m/s) Acceleration (m/s2)
Horizontal Axis $x={x}_{0}+v\cdot t$ ${v}_{x}={v}_{0x}=\text{cte}$ ${a}_{x}=0$
Vertical Axis $y=\mathrm{H}-\frac{1}{2}g{t}^{2}$ ${v}_{y}=-g\cdot t$ ${a}_{y}=-g$

Experiment and Learn

Data
g = 9.8 m/s2 |   |   |

Horizontal launched projectile

The blue ball in the figure represents a body suspended above the ground. You can drag it up to the initial height H that you want and select the initial velocity (v0) with which it will be horizontally launched. The gray line represents the trajectory that it will describe with your selected values.

Then press the Play button. Drag the time and observe how its position (x and y) and its velocity (vx and vy) are calculated at every moment of its descent toward the ground.

Verify that the projection of the body in the y-axis (green) experiences free fall motion and on the x-axis (red) it describes a uniform rectilinear motion.

### Equation of position and trajectory in horizontal launched projectile

The equation of position of a body helps us to determine at what point it is in each moment in time. In the case of a body that is moving in two dimensions, remember that, generically, displacement is described by:

$\stackrel{\to }{r}\left(t\right)=x\left(t\right)\stackrel{\to }{i}+y\left(t\right)\stackrel{\to }{j}$

Substituting in the previous expressions of the position in the horizontal axis (u.r.m.) and on the vertical axis (u.a.r.m.) in the generic equation of position, we can get to the expression of the equation of position for a horizontal launched projectile.

The equation of position for a horizontal launched projectile is given by:

$\stackrel{\to }{r}=\left({x}_{0}+v\cdot t\right)·\stackrel{\to }{i}+\left({y}_{0}-\frac{1}{2}·g·{t}^{2}\right)·\stackrel{\to }{j}$

On the other hand, to determine the trajectory that the body follows, i.e., its trajectory equation, we can combine the previous equations to eliminate t, leaving:

$y={y}_{0}-\frac{1}{2·{{v}_{0}}^{2}}·g·{x}^{2}={y}_{0}-k·{x}^{2}$

Where $k=\frac{1}{2·{{v}_{0}}^{2}}·g$  is constant throughout the trajectory.

## Solved exercises worksheet

Here you can test what you have learned in this section.

#### Initial height in horizontal launch

difficulty

At what height, should you place a cannon capable of launching projectiles at an initial velocity of 230 km/h on the horizontal axis if you would like the projectiles to fall at a distance of 250 meters from where they are shot?

#### Tennis and the horizontal launch

difficulty

A tennis ball located 2 m height is hit by a player with his racket. The ball shoots out horizontally with a velocity of 30 m/s. Answer the following questions:

a) How long does the ball takes to hit the ground?
b) What is the angle of the velocity vector with the x-axis at the time the ball reaches the ground?
c) If before being hit, the ball is 5 meters away from the net, how high does the ball pass above the net?

#### Horizontal launch of a rolling ball

difficulty

A golf ball rolls with constant velocity on the surface of a table 2.5 m above the ground. When it reaches the edge, it falls off the table as if horizontally launched so that at 0.4 s it is at a horizontal distance of 1 m from the edge of the table. Determine:

a) What was the constant velocity of the ball while rolling on the table?
b) Would you know how to determine how far horizontally will be the ball when it hits the ground?
c) What is the distance of the ball from the ground at 0.4 s?

#### Initial distance in horizontal launch

difficulty

Quarantine was declared in a passenger cruise ship due to a contagious viral intoxication. To help, the Red Cross sent a helicopter with a box full of drugs. Since the crew of the helicopter could not land on the ship, it is decided to drop the package on a mat on the ship deck.

Assuming that the cruise travels at 72 km/h and that the helicopter travels in the same direction at 108 km/h, at an altitude of 40 m, at what horizontal distance from the ship should the package be dropped? And at what distance if they are traveling toward each other?

## Formulas worksheet

Here is a full list of formulas for the section Horizontal Launch. By understanding each equation, you will be able to solve any problem that you may encounter at this level.

Click on the icon   to export them to any compatible external program.

#### Equation of position in rectilinear uniform motion -x-axis

$x={x}_{0}+v\cdot t$

#### Position equation in free fall

$y=\mathrm{H}-\frac{1}{2}g{t}^{2}$

#### Equation of speed in free fall

$v=-g\cdot t$

#### Equation of acceleration on the Earth surface

$a=-g$

#### Equation of acceleration in uniform rectilinear motion

$a=0$

#### Equation of velocity in uniform rectilinear motion

$v={v}_{0}=\text{cte}$