Among all the constant acceleration motions, or uniformly accelerated rectilinear movements (u.a.r.m.), there are two of particular interest: free fall and vertical launch. In this section, we will study vertical launch. Both are governed by the equations of the uniformly accelerated rectilinear motion (u.a.r.m.):

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

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

$a=\text{cte}$

Vertical launch

In vertical launch, an object is launched vertically up or down from a height H without taking into consideration any kind of friction with the air or any other obstacle. It is a uniformly accelerated rectilinear motion (u.a.r.m.) in which the acceleration is gravity. On the surface of the Earth, the acceleration of gravity can be considered constant and directed downward. It is designated by the letter g, and its value is 9.8 m/s2.

To study vertical launch motion, normally, we will use a system of reference whose origin of coordinates is located at the base of the positive y-axis, as can be seen in the picture.

Vertical launch upward

The body is launched upward from a height H, with a velocity greater than 0. As it ascends, its speed decreases until it reaches 0 (maximum height). From that point onwards, its velocity is negative and it begins to descend.

Vertical launch down

The body is launched downwards from a height H, with a speed smaller than 0 which will remain negative throughout the entire motion.

The vertical launch is a uniformly accelerated rectilinear motion (u.a.r.m.) or constant acceleration motion in which a body is launched vertically with some initial velocity from a certain height and do not find any resistance on its way. We can distinguish two cases based on the system of reference considered:

• We launch the body upward and therefore the initial velocity is positive (v0>0). In this case the equations for the upward vertical launch are:

$y=\mathrm{H}+{v}_{0}t-\frac{1}{2}g{t}^{2}$

$v={v}_{0}-g\cdot t$

$a=-g$

• We launch the body downward and therefore the initial velocity is negative (v0<0). In this case the equations for the downward vertical launch are:

$y=\mathrm{H}-{v}_{0}t-\frac{1}{2}g{t}^{2}$

$v=-{v}_{0}-g\cdot t$

$a=-g$

Where:

• y: Final position of the body. Its unit in the International System (SI) is the meter (m)
• v, v0: The final and initial velocity of the body respectively. Its unit in the International System (SI) is the meter per second (m/s)
• a: Acceleration of the body while in motion. Its unit in the International System (SI) is the meter per second squared (m/s2)
• t: Time spent on the motion. Its unit in the International System (SI) is the second (s)
• H: Height from which the body is launched. It is a measurement of length and therefore is unit is the meter (m)
• g: Value of the gravitational acceleration which on Earth surface can be considered equal to 9.8 m/s2

Experiment and Learn

Data
g = 9.8 m/s2 |   |

Vertical launch

The blue ball in the figure represents a body suspended above the ground. Drag it to the initial height H that you want and select the value of the initial velocity ( v) with which the body will be launched vertically. Then press the Play button to drop it.

Notice that, once the simulation is started, you can slide the time t(s) and see how, under the label Data, the corresponding values of position (y) and velocity (v) are calculated, as the body falls to the ground.

Verify that:

• If v0 is positive the body ascend to reach the highest point to then descend
• If v0 is negative the body descends from the start
• If the value of v0 is 0 it is a free fall motion

Solved exercises worksheet

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

A rookie equilibrist

difficulty

A rookie equilibrist is standing on a platform 12 meters above the ground. While practicing juggling with 2 balls, he stumbles and throws both balls at 9 m/s, however, he throws one up which we will call A and the other one down which we will call B. Considering that gravity is 10 m/s2, calculate

a) The time they are in the air.
b) Their velocity when they hit the ground.
c) The maximum height that ball A reaches.

A launch of negligible mass

difficulty

From a height of 40 meters an object of negligible mass is thrown downward with a velocity of 20 m/s. How long will it take to hit the ground? What will be its velocity on impact?

Vertical launch and free fall

difficulty

A stone is let to free fall to the bottom of a cliff with a height of 80 m. A second later a second stone is thrown downward so that it reaches the bottom at the same time as the first one

1. What was the launch velocity of the second stone?
2. What was the velocity of the first stone when they both hit the bottom?
3. How long was the second stone in the air?

Formulas worksheet

Here is a full list of formulas for the section Vertical 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 of downward vertical launch

$y=\mathrm{H}-{v}_{0}t-\frac{1}{2}g{t}^{2}$

Equation of position in upward vertical launch

$y=\mathrm{H}+{v}_{0}t-\frac{1}{2}g{t}^{2}$

Position equation of uniformly accelerated rectilinear motion - y-axis

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

Equation of velocity of downward vertical launch vertical

$v=-{v}_{0}-g\cdot t$

Equation of velocity of the upward vertical launch

$v={v}_{0}-g\cdot t$

Equation of acceleration on the Earth surface

$a=-g$