Questions.
1. Does gravity act on a body thrown up during its rise?
The force of gravity acts on all bodies, regardless of whether it is thrown up or at rest.
2. With what acceleration does a body thrown up move in the absence of friction? How does the speed of the body change in this case?
3. What determines the maximum lifting height of a body thrown up in the case when air resistance can be neglected?
The lifting height depends on the initial speed. (See previous question for calculations).
4. What can be said about the signs of the projections of the vectors of the instantaneous velocity of the body and the acceleration of free fall during the free movement of this body upwards?
When the body moves freely upwards, the signs of the projections of the velocity and acceleration vectors are opposite.
5. How were the experiments shown in Figure 30 carried out, and what conclusion follows from them?
For a description of the experiments, see pages 58-59. Conclusion: If only gravity acts on the body, then its weight is zero, i.e. it is in a state of weightlessness.
Exercises.
1. A tennis ball is thrown vertically upwards with an initial velocity of 9.8 m/s. How long will it take for the ball to rise to zero speed? How much movement from the place of the throw will the ball make in this case?
As we already know, gravity acts on all bodies that are on the surface of the Earth and near it. It doesn't matter if they are at rest or moving.
If a certain body falls freely to the Earth, then at the same time it will make uniformly accelerated motion, and the speed will increase constantly, since the velocity vector and the free fall acceleration vector will be co-directed with each other.
The essence of the movement vertically upwards
If we toss a body vertically upwards, and at the same time, we assume that there is no air resistance, then we can assume that it also makes uniformly accelerated motion, with free fall acceleration, which is caused by gravity. Only in this case, the speed that we gave the body during the throw will be directed upwards, and the acceleration of free fall is directed downwards, that is, they will be directed oppositely to each other. Therefore, the speed will gradually decrease.
After some time, the moment will come when the speed will be equal to zero. At this point, the body will reach its maximum height and stop for a moment. It is obvious that the greater the initial speed we give to the body, the greater the height it will rise by the time it stops.
- Further, the body will begin to fall down with uniform acceleration, under the influence of gravity.
How to solve problems
When you come across tasks for the movement of the body upwards, which does not take into account air resistance and other forces, but it is believed that only gravity acts on the body, then since the movement is uniformly accelerated, you can apply the same formulas as with a straight line uniformly accelerated motion with some initial speed V0.
Since in this case acceleration ax is the free fall acceleration of the body, then ax is replaced by gx.
- Vx=V0x+gx*t,
- Sx=V(0x)*t+(gx*t^2)/2.
It should also be taken into account that when moving up, the gravitational acceleration vector is directed downwards, and the velocity vector is upwards, that is, they are oppositely directed, and therefore their projections will have different signs.
For example, if the Ox axis is directed upwards, then the projection of the velocity vector when moving upwards will be positive, and the projection of the gravitational acceleration will be negative. This must be taken into account when substituting values into formulas, otherwise a completely wrong result will be obtained.
1588. How to determine the acceleration of free fall, having at its disposal a stopwatch, a steel ball and a scale up to 3 m high?
1589. What is the depth of the shaft if a stone freely falling into it reaches the bottom 2 s after the fall begins.
1590. The height of the Ostankino television tower is 532 m. A brick was dropped from its highest point. How long will it take him to hit the ground? Air resistance is ignored.
1591. The building of the Moscow state university on Sparrow Hills has a height of 240 m. A piece of facing has come off from the upper part of its spire and is falling freely down. How long will it take to reach the ground? Air resistance is ignored.
1592. A stone falls freely from a cliff. What distance will it cover in the eighth second from the beginning of the fall?
1593. A brick falls freely from the roof of a building 122.5 m high. What distance will the brick travel in the last second of its fall?
1594. Determine the depth of the well if the stone that fell into it touched the bottom of the well after 1 s.
1595. A pencil falls from a table 80 cm high to the floor. Determine the fall time.
1596. A body falls from a height of 30 m. What distance does it travel during the last second of its fall?
1597. Two bodies fall from different heights but reach the ground at the same time; in this case, the first body falls for 1 s, and the second - for 2 s. How far from the ground was the second body when the first began to fall?
1598. Prove that the time during which a body moving vertically upwards reaches greatest height h is equal to the time during which the body falls from this height.
1599. A body moves vertically downwards with an initial velocity. What are the simplest movements that can be decomposed into such a movement of the body? Write formulas for the speed and distance traveled for this movement.
1600. A body is thrown vertically upward at a speed of 40 m/s. Calculate at what height the body will be after 2 s, 6 s, 8 s and 9 s, counting from the beginning of the movement. Explain answers. To simplify the calculations, take g equal to 10 m/s2.
1601. With what speed must a body be thrown vertically upwards so that it comes back in 10 s?
1602. An arrow is launched vertically upwards with an initial speed of 40 m/s. In how many seconds will it fall back to the ground? To simplify the calculations, take g equal to 10 m/s2.
1603. The balloon rises vertically upwards uniformly at a speed of 4 m/s. A load is suspended from a rope. At an altitude of 217 m, the rope breaks. How many seconds will it take for the weight to hit the ground? Take g equal to 10 m/s2.
1604. A stone is thrown vertically upwards with an initial speed of 30 m/s. 3 s after the start of the movement of the first stone, the second stone was also thrown upwards with an initial speed of 45 m/s. At what height will the stones meet? Take g = 10 m/s2. Ignore air resistance.
1605. A cyclist climbs up a slope 100 m long. The speed at the beginning of the ascent is 18 km / h, and at the end 3 m / s. Assuming the movement is uniformly slow, determine how long the ascent took.
1606. Sledges move down the mountain with uniform acceleration with an acceleration of 0.8 m/s2. The length of the mountain is 40 m. Having rolled down the mountain, the sled continues to move uniformly and stops after 8 s ....
The patterns of falling bodies were discovered by Galileo Galilei.
The famous experiment with throwing balls from the Leaning Tower of Pisa (Fig. 7.1, a) confirmed his assumption that if air resistance can be neglected, then all bodies fall equally. When a bullet and a cannonball were thrown from this tower at the same time, they fell almost simultaneously (Fig. 7.1, b).
The fall of bodies in conditions where air resistance can be neglected is called free fall.
Let's put experience
The free fall of bodies can be observed using the so-called Newton's tube. Put a metal ball and a feather into a glass tube. Turning the tube over, we will see that the feather falls more slowly than the ball (Fig. 7.2, a). But if you pump air out of the tube, then the ball and feather will fall at the same speed (Fig. 7.2, b). 
This means that the difference in their fall in a tube with air is due only to the fact that air resistance for a feather plays a big role.
Galileo found that in free fall a body moves with constant acceleration, It is called the acceleration of free fall and is denoted by . It is directed downward and, as measurements show, is equal in modulus to approximately 9.8 m/s 2 . (At various points earth's surface g values vary slightly (within 0.5%).)
From the basic school physics course, you already know that the acceleration of bodies when they fall is due to the action of gravity.
When solving problems of a school course in physics (including USE assignments), g = 10 m/s 2 is accepted for simplification. Further, we will also do the same, without stipulating this in particular.
Consider first the free fall of a body without initial velocity.
In this and the following paragraphs, we will also consider the motion of a body thrown vertically upward and at an angle to the horizon. Therefore, we immediately introduce a coordinate system suitable for all these cases.
Let's point the x-axis horizontally to the right (we won't need it for now in this section), and the y-axis vertically up (Fig. 7.3). We choose the origin of coordinates on the surface of the earth. Let h denote the initial height of the body. 
A freely falling body moves with an acceleration, and therefore, with an initial velocity equal to zero, the velocity of the body at time t is expressed by the formula
1. Prove that the dependence of the velocity modulus on time is expressed by the formula
From this formula it follows that the speed of a freely falling body increases by about 10 m/s every second.
2. Plot v y (t) and v(t) for the first four seconds of the body falling.
3. A freely falling body without an initial velocity fell to the ground at a speed of 40 m/s. How long did the fall last?
From the formulas for uniformly accelerated motion without initial velocity it follows that
s y = g y t 2 /2. (3)
From here, for the displacement module we get:
s = gt 2 /2. (4)
4. How is the path traveled by the body related to the displacement modulus if the body falls freely without initial velocity?
5. Find the distance traveled by a freely falling body without initial velocity in 1 s, 2 s, 3 s, 4 s. Remember these path meanings: they will help you verbally solve many problems.
6. Using the results of the previous task, find the paths traversed by a freely falling body in the first, second, third and fourth seconds of the fall. Divide the found paths by five. Do you notice a simple pattern?
7. Prove that the dependence of the y coordinate of the body on time is expressed by the formula
y \u003d h - gt 2 / 2. (5)
Clue. Use formula (7) from § 6. Movement with rectilinear uniformly accelerated motion and the fact that the initial coordinate of the body is h, and the initial speed of the body is zero.
Figure 7.4 shows an example of a plot of y(t) for a freely falling body until it hits the ground. 
8. Using figure 7.4, check your answers to tasks 5 and 6.
9. Prove that the fall time of the body is expressed by the formula
Clue. Take advantage of the fact that at the moment of falling to the ground, the y-coordinate of the body is zero.
10. Prove that the modulus of the final velocity of the body vк (immediately before falling to the ground)
Clue. Use formulas (2) and (6).
11. What would be the speed of drops falling from a height of 2 km if the air resistance for them could be neglected, that is, they would fall freely?
The answer to this question will surprise you. Rain from such "droplets" would be destructive, not life-giving. Fortunately, the atmosphere saves us all: due to air resistance, the speed of raindrops at the earth's surface does not exceed 7–8 m/s.
2. Movement of a body thrown vertically upwards
Let a body be thrown from the surface of the earth vertically upwards with an initial velocity of 0 (Fig. 7.5). 
The velocity v_vec of the body at time t is expressed in vector form by the formula
In projections on the y-axis:
v y \u003d v 0 - gt. (9)
Figure 7.6 shows an example of a plot of v y (t) before the body falls to the ground. 
12. Determine from graph 7.6 at what point in time the body was at the top of the trajectory. What other information can be gleaned from this graph?
13. Prove that the time of lifting the body to the top of the trajectory can be expressed by the formula
t under = v 0 /g. (10)
Clue. Take advantage of the fact that at the top of the trajectory the speed of the body is zero.
14. Prove that the dependence of the coordinates of the body on time is expressed by the formula
y \u003d v 0 t - gt 2 /2. (eleven)
Clue. Use formula (7) from § 6. Displacement during rectilinear uniformly accelerated motion.
15. Figure 7.7 shows a plot of y(t). Find two different times when the body was at the same height and the time when the body was at the top of the trajectory. Have you noticed any pattern?
16. Prove that the maximum lifting height h is expressed by the formula
h = v 0 2 /2g (12)
Clue. Use formulas (10) and (11) or formula (9) from § 6. Displacement with rectilinear uniformly accelerated motion.
17. Prove that the final speed of a body thrown vertically upwards (that is, the speed of the body just before it hits the ground) is equal to but the modulus of its initial speed:
v k \u003d v 0. (13)
Clue. Use formulas (7) and (12).
18. Prove that the time of the entire flight
t floor = 2v 0 /g. (14)
Clue. Take advantage of the fact that at the moment of falling to the ground, the y-coordinate of the body becomes equal to zero.
19. Prove that
t floor = 2t under. (15)
Clue. Compare formulas (10) and (14).
Therefore, the rise of the body to the top of the trajectory takes the same time as the subsequent fall.
So, if air resistance can be neglected, then the flight of a body thrown vertically upwards naturally breaks down into two stages, which take the same time, an upward movement and a subsequent fall down to the starting point.
Each of these stages is, as it were, another stage “reversed in time”. Therefore, if we film the rise of a body thrown up to the top point on a video camera, and then we show the frames of this video in reverse order, then the audience will be sure that they are watching the fall of the body. And vice versa: the fall of the body shown in reverse order will look exactly like the rise of the body thrown vertically upwards.
This technique is used in the cinema: they film, for example, an artist who jumps from a height of 2–3 m, and then they show this filming in reverse order. And we admire the hero who easily takes off to a height unattainable for record holders.
Using the described symmetry between the ascent and descent of a body thrown vertically upwards, you will be able to perform the following tasks verbally. It is also useful to remember what the paths traversed by a freely falling body are equal to (task 4).
20. What is the distance traveled by a body thrown vertically upwards during the last second of ascent?
21. A body thrown vertically upwards has been at a height of 40 m twice with an interval of 2 s.
a) What is the maximum lift height?
b) What is the initial velocity of the body?
Additional questions and tasks
(All of the problems in this section assume that air resistance can be neglected.)
22. A body falls without initial velocity from a height of 45 m.
a) How long does the fall last?
b) What is the distance traveled by the body in the second second?
c) What is the distance traveled by the body in the last second of motion?
d) What is the final velocity of the body?
23. A body falls without initial velocity from a certain height within 2.5 s.
a) What is the final velocity of the body?
b) From what height did the body fall?
c) What is the distance traveled by the body in the last second of motion?
24. From the roof high house two drops fell with an interval of 1 s.
a) What is the speed of the first drop at the moment when the second drop comes off?
b) What is the distance between the drops at this moment?
c) What is the distance between the drops 2 s after the second drop begins to fall?
25. During the last τ seconds of falling without an initial velocity, the body has flown a distance l. Let us denote the initial height of the body h, the fall time t.
a) Express h in terms of g and t.
b) Express h - l in terms of g and t - τ.
c) From the resulting system of equations, express h in terms of l, g and τ.
d) Find the value of h at l = 30 m, τ = 1 s.
26. A blue ball is thrown vertically upwards with an initial velocity v0. At the moment when it reached the highest point, a red ball was thrown from the same starting point with the same initial speed.
a) How long did it take for the blue balloon to rise?
b) What is the maximum height of the blue ball?
c) How long after throwing the red ball did it collide with the moving blue one?
d) At what height did the balls collide?
27. A bolt came off from the ceiling of an elevator rising uniformly at a speed vl. Elevator cabin height h.
a) In what frame of reference is it more convenient to consider the movement of the bolt?
b) How long will the bolt fall?
c) What is the speed of the bolt just before it touches the floor: relative to the elevator? relative to the ground?
This video tutorial is for self-study topic "Movement of a body thrown vertically upwards". During this lesson, students will gain an understanding of the motion of a body in free fall. The teacher will talk about the movement of a body thrown vertically upwards.
In the previous lesson, we considered the issue of the motion of a body that was in free fall. Recall that we call free fall (Fig. 1) such a movement that occurs under the action of gravity. The force of gravity is directed vertically downward along the radius towards the center of the Earth, acceleration of gravity while equal to .
Rice. 1. Free fall
How will the movement of a body thrown vertically upwards differ? It will differ in that the initial velocity will be directed vertically upwards, i.e., it can also be considered along the radius, but not towards the center of the Earth, but, on the contrary, from the center of the Earth upwards (Fig. 2). But the acceleration of free fall, as you know, is directed vertically downwards. So, we can say the following: the movement of the body vertically upwards in the first part of the path will be slow motion, and this slow motion will also occur with free fall acceleration and also under the action of gravity.
Rice. 2 Movement of a body thrown vertically upwards
Let's turn to the figure and see how the vectors are directed and how it fits with the frame of reference.
Rice. 3. Movement of a body thrown vertically upwards
In this case, the reference system is connected to the earth. Axis Oy is directed vertically upwards, as is the initial velocity vector. The downward force of gravity acts on the body, which imparts to the body the acceleration of free fall, which will also be directed downward.
You can note next thing: the body will move slow, will rise to a certain height, and then will start rapidly fall down.
We have designated the maximum height, while .
The movement of a body thrown vertically upwards occurs near the surface of the Earth, when the free fall acceleration can be considered constant (Fig. 4).
Rice. 4. Near the surface of the Earth
Let us turn to the equations that make it possible to determine the speed, instantaneous speed and distance traveled during the considered movement. The first equation is the velocity equation: . The second equation is the equation of motion for uniformly accelerated motion: .
Rice. 5. Axis Oy pointing up
Consider the first frame of reference - the frame of reference associated with the Earth, the axis Oy directed vertically upwards (Fig. 5). The initial velocity is also directed vertically upwards. In the previous lesson, we already said that the acceleration of free fall is directed downward along the radius towards the center of the Earth. So, if we now reduce the velocity equation to a given reference frame, then we get the following: .
It is a projection of the speed at a certain point in time. The equation of motion in this case is: 
.
Rice. 6. Axis Oy pointing down
Consider another reference system, when the axis Oy directed vertically downwards (Fig. 6). What will change from this?
. The projection of the initial velocity will be with a minus sign, since its vector is directed upwards, and the axis of the selected reference system is directed downwards. In this case, the acceleration of free fall will be with a plus sign, because it is directed downwards. Motion equation: 
.
Another very important concept to consider is the concept of weightlessness.
Definition.Weightlessness- a state in which the body moves only under the influence of gravity.
Definition. Weight- the force with which the body acts on the support or suspension due to attraction to the Earth.
Rice. 7 Illustration for weight determination
If a body near the Earth or at a short distance from the Earth's surface moves only under the action of gravity, then it will not act on the support or suspension. This state is called weightlessness. Very often, weightlessness is confused with the concept of the absence of gravity. In this case, it must be remembered that weight is the action on the support, and weightlessness- this is when there is no effect on the support. Gravity is a force that always acts near the surface of the Earth. This force is the result of gravitational interaction with the Earth.
Let's take a look at one more important point associated with the free fall of bodies and the movement vertically upwards. When the body moves up and moves with acceleration (Fig. 8), an action occurs, leading to the fact that the force with which the body acts on the support exceeds the force of gravity. If this happens, this state of the body is called overload, or the body itself is said to be overloaded.
Rice. 8. Overload
Conclusion
The state of weightlessness, the state of overload - these are extreme cases. Basically, when a body is moving on a horizontal surface, the weight of the body and the force of gravity most often remain equal to each other.
Bibliography
- Kikoin I.K., Kikoin A.K. Physics: Proc. for 9 cells. avg. school - M.: Enlightenment, 1992. - 191 p.
- Sivukhin D.V. General course physics. - M .: State publishing house of technical
- theoretical literature, 2005. - T. 1. Mechanics. - S. 372.
- Sokolovich Yu.A., Bogdanova G.S. Physics: Handbook with examples of problem solving. - 2nd edition, redistribution. - X .: Vesta: Publishing house "Ranok", 2005. - 464 p.
- Internet portal "eduspb.com" ()
- Internet portal "physbook.ru" ()
- Internet portal "phscs.ru" ()
Homework
