VERTICAL SHOT: FORMULA, EQUATIONS, EXAMPLES - PHYSICAL - 2024

The vertical shot is a movement that takes place under the action of a force field, commonly that of gravity, and can be upward or downward. It is also known by the name of vertical launch.

The most immediate example is throwing up (or down if you prefer) a ball with the hand, of course, making sure to do it in a vertical direction. Disregarding air resistance, the motion that the ball follows fits perfectly with the Uniformly Varied Rectilinear Motion (MRUV) model.

Figure 1. Throwing a ball vertically upward is a good example of a vertical throw. Source: Pexels.

The vertical shot is a movement widely studied in introductory physics courses, as it is a sample of movement in one dimension, a very simple and useful model.

This model can not only be used to study the kinematics of objects under the action of gravity, but also, as will be seen later, describes the motion of particles in the midst of a uniform electric field.

Formulas and equations

The first thing you need is a coordinate system to mark the origin and label it with a letter, which in the case of vertical movements is the letter "y".

Next, the positive direction + y is selected, which is generally upwards, and the –y direction is usually taken downwards (see figure 2). All this unless the problem solver decides otherwise, since another option is to take the direction of the movement as positive, whatever it may be.

Figure 2. Usual sign convention in vertical shooting. Source: F. Zapata.

In any case, it is recommended that the origin coincide with the launch point and _or, because in this way the equations are simplified, although any desired position can be taken to start studying the movement.

Vertical throw equations

Once the coordinate system and the origin are established, we go to the equations. The magnitudes that describe the movement are:

-Initial speed v _o

-Acceleration to

-Speed v

-Initial position x _o

-Position x

-Displacement D x

-Time t

All except time are vectors, but since it is a one-dimensional movement with a certain direction, what matters then is to use + or - signs to indicate where the magnitude in question is going. In the case of vertical draft, gravity always goes downwards and, unless otherwise specified, it is assigned a sign -.

Following are the equations adapted for vertical draft, substituting “x” for “y” and “a” for “g”. In addition, the sign (-) corresponding to gravity directed downwards will be included at once:

1) Position: y = y _o + v _o.t - ½ gt ²

2) Velocity: v = v _o - gt

3) Velocity as a function of displacement Δ y: v ² = v _o ² - 2.g. Δ and

Examples

Below are application examples for vertical shooting. In its resolution, the following must be taken into account:

- "g" has a constant value that on average is 9.8 m / s ² or approximately 10 m / s ² if preferred to facilitate calculations when too much precision is not required.

-When v _o is 0, these equations are reduced to those of free fall.

-If the launch is upwards, the object needs to have an initial speed that allows it to move. Once in motion, the object reaches a maximum height that will depend on how great the initial velocity is. Of course, the higher the altitude, the more time the mobile will spend in the air.

-The object returns to the starting point with the same speed with which it was thrown, but the speed is directed downwards.

-For a vertical downward launch, the higher the initial velocity, the sooner the object will hit the ground. Here the distance traveled is set according to the height selected for the launch.

-In the vertical shot upwards, the time it takes for the mobile to reach the maximum height is calculated by making v = 0 in equation 2) of the previous section. This is the maximum time t _max:

-The maximum height and _max is cleared from equation 3) of the previous section by also making v = 0:

If y _o = 0, it reduces to:

Worked example 1

A ball with v _o = 14 m / s is thrown vertically upward from the top of an 18 m high building. The ball is allowed to continue its way down to the sidewalk. Calculate:

a) The maximum height reached by the ball with respect to the ground.

b) The time it was in the air (flight time).

Figure 3. A ball is thrown vertically upward from the roof of a building. Source: F. Zapata.

Solution

The figure shows the raising and lowering movements of the ball separately for clarity, but both occur along the same line. The initial position is taken at y = 0, so the final position is y = - 18 m.

a) The maximum height measured from the roof of the building is y _max = v _or ² / 2g and from the statement it is read that the initial velocity is +14 m / s, then:

Substituting:

It is an equation of the second degree that is easily solved with the help of a scientific calculator or using the solver. The solutions are: 3.82 and -0.96. The negative solution is discarded since, being a time, it lacks physical sense.

The flight time of the ball is 3.82 seconds.

Worked example 2

A positively charged particle with q = +1.2 millicoulombs (mC) and mass m = 2.3 x 10 ^-10 Kg is projected vertically upwards, starting from the position shown in the figure and with initial velocity v _o = 30 km / s.

Between the charged plates there is a uniform electric field E, directed vertically downwards and with a magnitude of 780 N / C. If the distance between the plates is 18 cm, will the particle collide with the top plate? Neglect the gravitational attraction on the particle, as it is extremely light.

Figure 4. A positively charged particle moves in a manner similar to a ball thrown vertically upward, when it is immersed in the electric field in the figure. Source: modified by F. Zapata from Wikimedia Commons.

Solution

In this problem the electric field E is the one that produces a force F and the consequent acceleration. Being positively charged, the particle is always attracted to the lower plate, however when it is projected vertically upwards it will reach a maximum height and then return to the lower plate, just like the ball in the previous examples.

By definition of electric field:

You need to use this equivalency before substituting values:

Thus the acceleration is:

For the maximum height, the formula from the previous section is used, but instead of using “g”, this acceleration value is used:

and _max = v _or ² / 2a = (30,000 m / s) ² /2 x 4.07 X 10 ⁹ m / s ² = 0.11 m = 11 cm

It does not collide with the upper plate, since this is 18 cm from the starting point, and the particle only reaches 11 cm.

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. Volume 1. 50 - 53.
4. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9 ^na Ed. Cengage Learning. 43 - 55.
5. Wilson, J. 2011. Physics 10. Pearson Education. 133-149.