Vertical shot formulas, equations, examples

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Simon Doyle

The vertical shot It is a movement that takes place under the action of a force field, commonly that of gravity, and can be ascending or descending. 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 perfectly fits 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 the 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..

Article index

  • 1 Formulas and equations
    • 1.1 Vertical throw equations
  • 2 Examples
    • 2.1 Worked example 1
    • 2.2 Worked example 2
  • 3 References

Formulas and equations

The first thing that is needed is a coordinate system to indicate the origin and label it with a letter, which in the case of vertical movements is the letter "Y".

Then the positive direction is selected +Y, which is usually upwards and the sense -Y which is usually taken down (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 coincides with the launch point. Yor, because in this way the equations are simplified, although you can take any position you want to start studying the motion.

Vertical throw equations

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

-Initial velocity vor

-Acceleration to

-Velocity v

-Initial position xor

-Position x

-Displacement Dx

-Weather 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" Y "to" for "g”. In addition, the sign (-) corresponding to gravity directed downwards will be included at once:

1) Position: y = yor + vor.t - ½ g.ttwo

2) Speed: v = vor - g.t

3) Speed ​​as a function of displacement ΔY: vtwo = vortwo - 2 g. ΔY

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 / stwo or about 10 m / stwo if preferred to facilitate calculations when too much precision is not required.

-When vor okay 0, these equations reduce to those of freefall.

-If the launch is upward, the object needs to have an initial velocity 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 downward vertical throw, 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 up, the time it takes for the mobile to reach the maximum height is calculated by doing v = 0 in equation 2) of the previous section. This is the maximum time tmax:

0 = vor - g. tmax tmax = vor / g

-The maximum height Ymax is cleared from equation 3) of the previous section by doing likewise v = 0:

0 = vortwo - 2 g. Δy ⇒ 0 = vortwo - 2 g. (Ymax - Yor) ⇒ andmax = andor  + vortwo / 2 g

Yes Yor = 0, It is reduced to:

Ymax = vortwo / 2 g

Worked example 1

A ball is thrown vertically upwards with vor = 14 m / s, 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 that the final position is y = - 18 m.

a) The maximum height measured from the roof of the building is Ymax = vortwo / 2 g and from the statement we can read that the initial velocity is +14 m / s, then:

Ymax = (14 m / s)two / 2 x 9.8 m / stwo = 10 m (Regarding the roof)

Hmax = 10 m + 18 m = 28 m (Regarding the sidewalk).

b) To find the total time or flight time the ball lasts in the air, the equation will be used y = yor + vor.t - ½ g.ttwo, with the following values ​​and signs:

y = - 18 m

Yor = 0 m

vor = +14 m / s

Substituting:

- 18 = 14.t - ½ 9.8 .ttwo

- 4.9 ttwo+14.t + 18 = 0 

4.9 ttwo-14.t - 18 = 0

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, since it is 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 vor = 30 km / s.

Between the charged plates there is an electric field AND uniform, directed vertically downwards and with 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 AND 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:

E = F / q = m.a / q ⇒ a = q.E / m

You need to use this equivalency before substituting values:

1 mC = 1 x 10-3 C

Thus the acceleration is:

a = 1.2 x 10-3 x 780 / 2.3 x 10 -10m / stwo = 4.07 x 10m / stwo

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

Ymax = vortwo / 2a = (30,000 m / s)two/ 2 x 4.07 x 10m / stwo = 0.11m = 11cm

It does not collide with the upper plate, since it 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. 6ta Shortened edition. Cengage Learning. 23 - 27.
  2. Rex, A. 2011. Fundamentals of Physics. Pearson. 33 - 36
  3. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1. 50 - 53.
  4. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9na Ed. Cengage Learning. 43 - 55.
  5. Wilson, J. 2011. Physics 10. Pearson Education. 133 - 149.

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