The patterns of falling bodies were discovered by Galileo Galilei.
The famous experiment with throwing balls from an inclined Leaning Tower of Pisa (Fig. 7.1, a) confirmed his assumption that if the 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 tube. Put a metal ball and a feather in 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 out air from the tube, then the ball and the feather will fall at the same speed (Fig. 7.2, b). 
This means that the difference in their fall in the tube with air is due only to the fact that the air resistance for the feather plays an important role.
Galileo established that during free fall, the body moves with constant acceleration, It is called acceleration free fall and denote. It is directed downwards and, as measurements show, is equal in magnitude to about 9.8 m / s 2. (At different points on the earth's surface, the g values differ slightly (within 0.5%).)
You already know from the physics course in basic school that the acceleration of bodies during a fall is due to the action of gravity.
When solving problems of a school physics course (including USE tasks), for simplicity, g = 10 m / s 2 is taken. Further, we will also do the same, without specifying it.
Consider first the free fall of the body without initial speed.
In this and the following paragraphs, we will also consider the movement 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 direct the x-axis horizontally to the right (we won't need it in this section yet), and the y-axis vertically upward (Fig. 7.3). We select the origin of coordinates on the surface of the earth. Let's denote by h the initial height of the body. 
A freely falling body moves with acceleration, and therefore, at equal to zero the initial velocity, 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
It follows from this formula that the speed of a freely falling body increases by about 10 m / s every second.
2. Draw graphs of dependence v y (t) and v (t) for the first four seconds of the fall of the body.
3. A body freely falling without 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 the initial velocity it follows that
s y = g y t 2/2. (3)
From here, for the movement module, we get:
s = gt 2/2. (4)
4. How is the path traversed by the body related to the module of displacement, if the body falls freely without initial velocity?
5. Find the value of the path traversed by a body freely falling 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 values of the found paths by five. Do you see a simple pattern?
7. Prove that the dependence of the y-coordinate of the body on time is expressed by the formula
y = h - gt 2/2. (5)
Prompt. Use the formula (7) from § 6. Displacement with rectilinear uniformly accelerated motion and the fact that the initial coordinate of the body is equal to h, and the initial velocity of the body is zero.
Figure 7.4 shows an example of a graph of the dependence y (t) for a freely falling body until it falls to 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
Prompt. Take advantage of the fact that at the moment of falling to the ground, the y-coordinate of the body is equal to zero.
10. Prove that the modulus of the final velocity of the body vk (immediately before falling to the ground)
Prompt. Use formulas (2) and (6).
11. What would the speed of drops falling from a height of 2 km be equal to if the air resistance for them could be neglected, that is, they would fall freely?
The answer to this question will surprise you. The 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. The movement of a body thrown vertically upward
Let the body be thrown from the surface of the earth vertically upward with an initial speed of 0 (Fig. 7.5). 
The velocity v_vec of the body at time t in vector form is expressed by the formula
In projections to the y-axis:
v y = v 0 - gt. (nine)
Figure 7.6 shows an example of a graph of dependence v y (t) until the moment the body falls to the ground. 
12. Determine according to graph 7.6, at what moment in time the body was at the top point of the trajectory. What other information can be extracted from this graph?
13. Prove that the lifting time of the body to the top point of the trajectory can be expressed by the formula
t under = v 0 / g. (ten)
Prompt. 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 = v 0 t - gt 2/2. (eleven)
Prompt. Use formula (7) from § 6. Displacement with uniformly accelerated rectilinear motion.
15. Figure 7.7 shows the graph of the dependence y (t). Find two different moments in time when the body was at the same height and the moment in 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)
Prompt. 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 upward (that is, the speed of the body immediately before falling to the ground) is equal but the modulus of its initial speed:
v k = v 0. (13)
Prompt. Use formulas (7) and (12).
18. Prove that the time of the entire flight
t floor = 2v 0 / g. (fourteen)
Prompt. 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 floor. (15)
Prompt. Compare formulas (10) and (14).
Consequently, lifting 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, taking the same time, - upward movement and subsequent fall down to the starting point.
Each of these stages represents, as it were, another stage “reversed in time”. Therefore, if we shoot with a video camera the rise of the body thrown up to the upper point, and then we show the frames of this video in the reverse order, then the viewers will be sure that they are observing the fall of the body. And vice versa: shown in reverse order, the fall of the body will look exactly like the lifting of a body thrown vertically upward.
This technique is used in films: for example, they shoot an artist who jumps from a height of 2-3 m, and then show this shooting in the reverse order. And we admire the hero who easily takes off to a height unattainable for record holders.
Using the described symmetry between lifting and lowering a body thrown vertically upward, you will be able to perform the following tasks orally. It is also useful to remember what the paths traversed by a freely falling body are equal (task 4).
20. What is the path taken by a body thrown vertically upward in the last second of ascent?
21. A body thrown vertically upward has been at a height of 40 m twice with an interval of 2 s.
a) What is the maximum body height?
b) What is the initial velocity of the body?
Additional questions and tasks
(Throughout this section, it is assumed that air resistance is negligible.)
22. The body falls without initial speed from a height of 45 m.
a) How long does the fall last?
b) What distance does the body fly in the second second?
c) What distance does the body fly in the last second of movement?
d) What is the final velocity of the body?
23. The body falls without initial speed from a certain height for 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 the body flew in the last second of the movement?
24.From the roof high house with an interval of 1 s, two drops fell.
a) What is the velocity of the first drop at the moment when the second drop is torn off?
b) What is the distance between the drops equal at this moment?
c) What is the distance between the drops 2 s after the start of the fall of the second drop?
25. During the last τ seconds of the fall without initial velocity, the body flew the distance l. Let us denote the initial height of the body as h, and the time of fall 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. The blue ball is thrown vertically upward with an initial velocity v0. The moment he reached highest point, a red ball is thrown from the same starting point with the same initial speed.
a) How long did the blue ball 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?
d) At what height did the balls collide?
27. A bolt came off the ceiling of the elevator, rising evenly with a speed vl. Elevator car height h.
a) In which 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 touching the floor: relative to the elevator? relative to the ground?
Let the body begin to fall freely from rest. In this case, the formulas of uniformly accelerated motion without initial velocity with acceleration are applicable to its motion. Let's designate the initial height of the body above the ground through, the time of its free fall from this height to the ground - through and the speed reached by the body at the moment of falling to the ground - through. According to the formulas in § 22, these quantities will be related by the relations
(54.1)
(54.2)
Depending on the nature of the problem, it is convenient to use one or the other of these relations.
Let us now consider the motion of a body, which is imparted with a certain initial velocity directed vertically upward. In this problem, it is convenient to assume that the upward direction is positive. Since the acceleration of gravity is directed downward, the movement will be equally slow with negative acceleration and with a positive initial velocity. The speed of this movement at the moment of time is expressed by the formula
and the lifting height at this moment above the starting point is by the formula
(54.5)
When the speed of the body decreases to zero, the body will reach its highest point of ascent; it will happen at the moment for which
After this moment, the speed will become negative and the body will begin to fall down. This means that the time for lifting the body
Substituting the lifting time into formula (54.5), we find the body's lifting height:
(54.8)
Further movement of the body can be considered as a fall without initial velocity (the case considered at the beginning of this section) from a height. Substituting this height into formula (54.3), we find that the speed that the body reaches at the moment of falling to the ground, i.e., returning to the point from where it was thrown up, will be equal to the initial speed of the body (but, of course, it will be directed in the opposite direction - way down). Finally, from formula (54.2) we conclude that the time of the body falling from the highest point is equal to the time of lifting the body to this point.
5 4.1. The body freely falls without initial speed from a height of 20 m. At what height will it reach a speed equal to half the speed at the moment of falling to the ground?
54.2. Show that a body thrown vertically upward passes each point of its trajectory with the same velocity modulus on the way up and on the way down.
54.3. Find the speed when a stone thrown from a tower of height hits the ground: a) without initial speed; b) with an initial speed directed vertically upward; c) with an initial speed directed vertically downward.
54.4. A stone thrown vertically upward flew past the window 1 s after being thrown on the way up and 3 s after being thrown on the way down. Find the height of the window above the ground and the initial velocity of the stone.
54.5. When firing vertically at air targets, the projectile fired from an anti-aircraft gun reached only half the distance to the target. A shell fired from another weapon reached its target. How many times is the muzzle velocity of the second gun greater than the velocity of the first?
54.6. What is the maximum height to which a stone thrown vertically upwards can rise if, after 1.5 s, its speed has halved?
The movement of a body thrown vertically upward
Level I. Read the text
If some body freely falls to the Earth, then it will perform 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.
If we throw some body vertically upwards, and at the same time assume that there is no air resistance, then we can assume that it also performs uniformly accelerated motion, with the acceleration of free fall, which is caused by the force of 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 downward, that is, they will be oppositely directed towards each other. Therefore, the speed will gradually decrease.
After a while, the moment will come when the speed will be zero. At this moment, the body will reach its maximum height and stop for a moment. Obviously, the greater the initial velocity we give to the body, the greater the height it will rise to the moment it stops.
All formulas for uniformly accelerated motion apply to the motion of a body thrown upwards. V0 always> 0
The movement of a body thrown vertically upwards is straight motion with constant acceleration. If you direct the coordinate axis OY vertically upward, aligning the origin with the Earth's surface, then to analyze free fall without initial velocity, you can use the formula https://pandia.ru/text/78/086/images/image002_13.gif "width =" 151 "height =" 57 src = ">
Near the Earth's surface, provided that there is no noticeable influence of the atmosphere, the speed of a body thrown vertically upward changes in time according to a linear law: https://pandia.ru/text/78/086/images/image004_7.gif "width =" 55 "height = "28">.
The speed of the body at a certain height h can be found by the formula:
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The height of the body in a certain time, knowing the final speed
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IIIlevel. Solve the problems. For 9 b. 9a solves from a problem book!
1. The ball is thrown vertically upward at a speed of 18 m / s. What movement will he make in 3 seconds?
2. An arrow fired from the bow vertically upward at a speed of 25 m / s hits the target after 2 s. What speed did the arrow have at the time the target was reached?
3. From a spring pistol, a ball was shot vertically upwards, which rose to a height of 4.9 m. At what speed did the ball fly out of the pistol?
4. The boy threw the ball vertically up and caught it after 2 s. How high did the ball rise and what is its initial velocity?
5. With what initial speed should the body be thrown vertically upward so that after 10 s it moves downward at a speed of 20 m / s?
6. “Humpty Dumpty was sitting on a wall (20 m high),
Humpty Dumpty collapsed in his sleep.
Do we need all the royal cavalry, all the royal men,
to Humpty, to Dumpty, Humpty Dumpty,
Dumpty-Humpty to collect "
(if it breaks only at 23 m / s?)
So is all the royal cavalry needed?
7. Now the thunder of sabers, spurs, sultan,
And the camera-junker caftan
Patterned - beauties are seduced,
Wasn't it a temptation
When from the guard, others from the court
They came here for a while!
The women were shouting: hurray!
And they threw their caps into the air.
"Woe from Wit."
The girl Catherine was throwing her cap up at a speed of 10 m / s. At the same time, she stood on the balcony of the 2nd floor (at a height of 5 meters). How long will the cap be in flight if it falls under the feet of the brave hussar Nikita Petrovich (naturally standing under the balcony on the street).
1588. How to determine the acceleration of gravity, having at its disposal a stopwatch, a steel ball and a scale up to 3 m high?
1589. What is the depth of the mine, if a stone falling freely into it reaches the bottom 2 s after the beginning of the fall.
1590. The height of the Ostankino TV tower is 532 m. A brick was dropped from its highest point. How long does it take for it to fall to the ground? Disregard air resistance.
1591. Building of Moscow state university on Vorobyovy Gory it has a height of 240 m. A piece of cladding has come off the top of its spire and falls freely downward. How long will it take to reach the ground? Disregard air resistance.
1592. The stone falls freely from the cliff. What path will he cover in the eighth second from the beginning of the fall?
1593. A brick falls freely from the roof of a building with a height of 122.5 m. What path will a brick take in the last second of its fall?
1594. Determine the depth of the well, if a 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 cover during the last second of its fall?
1597. Two bodies fall from different heights, but reach the ground at the same time; 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 highest height h is equal to the time during which the body falls from this height.
1599. The body moves vertically downward with an initial speed. What are the simplest movements that such a body movement can be decomposed into? Write formulas for the speed and distance traveled for this movement.
1600. The body is thrown vertically upward at a speed of 40 m / s. Calculate at what height the body will be in 2 s, 6 s, 8 s and 9 s, counting from the beginning of the movement. Explain the answers. To simplify calculations, take g equal to 10 m / s2.
1601. With what speed should the body be thrown vertically upwards so that it comes back in 10 s?
1602. The arrow is launched vertically upward with an initial speed of 40 m / s. In how many seconds will it fall back to the ground? To simplify calculations, take g equal to 10 m / s2.
1603. The balloon rises evenly vertically upward at a speed of 4 m / s. A load is suspended from it on a rope. At a height of 217 m, the rope breaks. In how many seconds will the load fall to the ground? Take g equal to 10 m / s2.
1604. The stone was thrown vertically upward with an initial speed of 30 m / s. 3 seconds after the start of the movement of the first stone, the second stone was also thrown upward with an initial speed of 45 m / s. At what height will the stones meet? Take g = 10 m / s2. Neglect air resistance.
1605. The cyclist climbs up a 100 m incline. The speed at the beginning of the ascent is 18 km / h, and at the end of 3 m / s. Assuming the movement is equally slow, determine how long the rise lasted.
1606. Sleds are moving 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 from the mountain, the sled continues to move with equal slowness and stops after 8 seconds….
You know that when any body falls to the Earth, its speed increases. Long time believed that the Earth imparts different accelerations to different bodies. Simple observations seem to confirm this.
But only Galileo was able to experimentally prove that in reality this is not so. Air resistance must be taken into account. It is this that distorts the picture of the free fall of bodies, which could be observed in the absence of earth's atmosphere... To test his assumption, Galileo, according to legend, observed the fall from the famous inclined Leaning Tower of Pisa of various bodies (cannonball, musket bullet, etc.). All these bodies reached the Earth's surface almost simultaneously.
The experiment with the so-called Newton tube is especially simple and convincing. Various objects are placed in a glass tube: pellets, pieces of cork, fluffs, etc. If you now turn the tube over so that these objects can fall, then a pellet will flash faster, followed by pieces of cork and, finally, a fluff will smoothly descend (Fig. 1, a). But if you pump air out of the tube, then everything will happen completely differently: the fluff will fall, keeping up with the pellet and the plug (Fig. 1, b). This means that its movement was delayed by air resistance, which had a lesser effect on the movement of, for example, traffic jams. When these bodies are affected only by the attraction to the Earth, then they all fall with the same acceleration.
Rice. 1
- Free fall is the movement of a body only under the influence of attraction to the Earth.(no air resistance).
Acceleration communicated to all bodies the globe are called acceleration of gravity... We will denote its module by the letter g... Free fall does not necessarily represent a downward movement. If the initial velocity is directed upward, then the body will fly upward for some time during free fall, decreasing its velocity, and only then will it begin to fall downward.
Body movement vertically
- Equation of projection of speed on an axis 0Y: $ \ upsilon _ (y) = \ upsilon _ (0y) + g_ (y) \ cdot t, $
the equation of motion along the axis 0Y: $ y = y_ (0) + \ upsilon _ (0y) \ cdot t + \ dfrac (g_ (y) \ cdot t ^ (2)) (2) = y_ (0) + \ dfrac (\ upsilon _ (y ) ^ (2) - \ upsilon _ (0y) ^ (2)) (2g_ (y)), $
where y 0 - starting coordinate of the body; υ y- the projection of the final speed on the axis 0 Y; υ 0 y- projection of the initial speed on the axis 0 Y; t- time during which the speed changes (s); g y- the projection of the acceleration of gravity on the axis 0 Y.
- If axis 0 Y point upwards (fig. 2), then g y = –g, and the equations take the form
Rice. 2 Hidden data When the body moves down
- "The body falls" or "the body fell" - υ 0 at = 0.
ground surface, then:
- "The body fell to the ground" - h = 0.
- "The body has reached its maximum height" - υ at = 0.
If we take as the starting point ground surface, then:
- "The body fell to the ground" - h = 0;
- "The body was thrown from the ground" - h 0 = 0.
- Rise time body to maximum height t under is equal to the time of falling from this height to the starting point t pad, and total time flight t = 2t under.
- The maximum lifting height of a body thrown vertically upwards from zero height (on maximum height υ y = 0)
The movement of a body thrown horizontally
A particular case of the movement of a body thrown at an angle to the horizon is the movement of a body thrown horizontally. The trajectory is a parabola with apex at the drop point (Fig. 3).
Rice. 3
Such a movement can be decomposed into two:
1) uniform traffic horizontally with a speed υ 0 NS (a x = 0)
- velocity projection equation: $ \ upsilon _ (x) = \ upsilon _ (0x) = \ upsilon _ (0) $;
- equation of motion: $ x = x_ (0) + \ upsilon _ (0x) \ cdot t $;
To describe the motion along the 0 axis Y the formulas for uniformly accelerated vertical motion are applied:
- velocity projection equation: $ \ upsilon _ (y) = \ upsilon _ (0y) + g_ (y) \ cdot t $;
- equation of motion: $ y = y_ (0) + \ dfrac (g_ (y) \ cdot t ^ (2)) (2) = y_ (0) + \ dfrac (\ upsilon _ (y) ^ (2)) (2g_ ( y)) $.
- If axis 0 Y point up then g y = –g, and the equations take the form:
- Range of flight defined by the formula: $ l = \ upsilon _ (0) \ cdot t_ (nad). $
- Body speed at any given time t will be equal (Fig. 4):
where υ NS = υ 0 x , υ y = g y t or υ NS= υ ∙ cos α, υ y= υ ∙ sin α.
Rice. 4
When solving free fall problems
1. Select the reference body, specify the start and end positions of the body, select the direction of the axes 0 Y and 0 NS.
2. Draw the body, indicate the direction of the initial velocity (if it is equal to zero, then the direction of the instantaneous velocity) and the direction of the gravitational acceleration.
3. Write down the original equations in projections on the 0 axis Y(and, if necessary, on axis 0 X)
$ \ begin (array) (c) (0Y: \; \; \; \; \; \ upsilon _ (y) = \ upsilon _ (0y) + g_ (y) \ cdot t, \; \; \; (1)) \\ () \\ (y = y_ (0) + \ upsilon _ (0y) \ cdot t + \ dfrac (g_ (y) \ cdot t ^ (2)) (2) = y_ (0) + \ dfrac (\ upsilon _ (y) ^ (2) - \ upsilon _ (0y) ^ (2)) (2g_ (y)), \; \; \; \; (2)) \\ () \ \ (0X: \; \; \; \; \; \ upsilon _ (x) = \ upsilon _ (0x) + g_ (x) \ cdot t, \; \; \; (3)) \\ () \\ (x = x_ (0) + \ upsilon _ (0x) \ cdot t + \ dfrac (g_ (x) \ cdot t ^ (2)) (2). \; \; \; (4)) \ end (array) $
4. Find the values of the projections of each quantity
x 0 = …, υ x = …, υ 0 x = …, g x = …, y 0 = …, υ y = …, υ 0 y = …, g y = ….Note... If axis 0 NS is directed horizontally, then g x = 0.
5. Substitute the obtained values into equations (1) - (4).
6. Solve the resulting system of equations.
Note... As you develop the skill of solving such problems, point 4 can be done in your mind, without writing in a notebook.