Movement of a body thrown vertically by a formula. Free fall of bodies

You know that when any body falls to the Earth, its speed increases. For a long time it was believed that the Earth imparts different accelerations to different bodies. Simple observations seem to confirm this.

But only Galileo managed to prove empirically that this is not the case in reality. Air resistance must be taken into account. It is it that distorts the picture of the free fall of bodies, which could be observed in the absence of the earth's atmosphere. To test his assumption, Galileo, according to legend, observed the fall of various bodies (cannonball, musket ball, etc.) from the famous Leaning Tower of Pisa. All these bodies reached the Earth's surface almost simultaneously.

The experiment with the so-called Newton's tube is especially simple and convincing. Various objects are placed in a glass tube: pellets, pieces of cork, fluff, etc. If we now turn the tube over so that these objects can fall, then the pellet will flash through the fastest, followed by pieces of cork, and, finally, the fluff will smoothly fall (Fig. 1a). But if you pump air out of the tube, then everything will happen in a completely different way: the fluff will fall, keeping up with the pellet and cork (Fig. 1, b). This means that its movement was delayed by air resistance, which to a lesser extent affected the movement, for example, of traffic jams. When only attraction to the Earth acts on these bodies, then they all fall with the same acceleration.

Rice. one

• Free fall is the movement of a body only under the influence of attraction to the Earth(without air resistance).

The acceleration imparted to all bodies by the globe is called free fall acceleration. We will denote its module by the letter g. Free fall does not necessarily represent downward movement. If the initial velocity is directed upwards, then the body in free fall will fly upwards for some time, reducing its speed, and only then will it begin to fall downwards.

Vertical body movement

• The equation for the projection of speed on the axis 0Y: $\upsilon _(y) =\upsilon _(0y) +g_(y) \cdot t,$

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 - initial coordinate of the body; υ y- projection of final speed on 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- projection of free fall acceleration on axis 0 Y.

• If axis 0 Y point upwards (Fig. 2), then g y = –g, and the equations take the form

$\begin(array)(c) (\upsilon _(y) =\upsilon _(0y) -g\cdot t,) \\ (\, y=y_(0) +\upsilon _(0y) \cdot t-\dfrac(g\cdot t^(2) )(2) =y_(0) -\dfrac(\upsilon _(y)^(2) -\upsilon _(0y)^(2) )(2g ) .) \end(array)$

Rice. 2 Hidden data When the body moves down

• "body falls" or "body fell" - υ 0 at = 0.

land surface, then:

• body fell to the ground h = 0.

When moving the body up

• "the body has reached its maximum height" - υ at = 0.

If we take as the origin land surface, then:

• 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 equal to the time of fall from this height to the starting point t fall, and the total flight time t = 2t under.

• The maximum lifting height of a body thrown vertically upwards from zero height (at the maximum height υ y = 0)

$h_(\max ) =\dfrac(\upsilon _(x)^(2) -\upsilon _(0y)^(2) )(-2g) =\dfrac(\upsilon _(0y)^(2) )(2g).$

Movement of a body thrown horizontally

A special case of the motion of a body thrown at an angle to the horizon is the motion of a body thrown horizontally. The trajectory is a parabola with a vertex at the throwing point (Fig. 3).

Rice. 3

This movement can be decomposed into two:

1) uniform motion horizontally with speed υ 0 X (a x = 0)

• velocity projection equation: $\upsilon _(x) =\upsilon _(0x) =\upsilon _(0) $;
• equation of motion: $x=x_(0) +\upsilon _(0x) \cdot t$;

2) uniformly accelerated motion vertically with acceleration g and initial speed υ 0 at = 0.

To describe the movement along the axis 0 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:

$\begin(array)(c) (\upsilon _(y) =-g\cdot t,\, ) \\ (y=y_(0) -\dfrac(g\cdot t^(2) )(2 ) =y_(0) -\dfrac(\upsilon _(y)^(2) )(2g) .) \end(array)$

• Range of flight is determined by the formula: $l=\upsilon _(0) \cdot t_(nad) .$

• The speed of the body at any given time t will be equal to (Fig. 4):

$\upsilon =\sqrt(\upsilon _(x)^(2) +\upsilon _(y)^(2) ) ,$

where v X = υ 0 x , υ y = g y t or υ X= υ∙cosα, υ y= υ∙sinα.

Rice. 4

When solving free fall problems

1. Select the reference body, specify the initial and final positions of the body, select the direction of the axes 0 Y and 0 X.

2. Draw a 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 free fall acceleration.

3. Write down the initial 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 X directed horizontally, then g x = 0.

5. Substitute the obtained values ​​into equations (1) - (4).

6. Solve the resulting system of equations.

Note. As the skill of solving such problems is developed, point 4 can be done in the mind, without writing in a notebook.

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?

The motion of a body thrown vertically upwards

I level. Read the text

If a certain body falls freely 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 toss some body vertically upwards, and at the same time 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 to 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.

All formulas for uniformly accelerated motion are applicable to the motion of a body thrown upwards. V0 always > 0

The motion of a body thrown vertically upwards is a rectilinear motion with constant acceleration. If you direct the OY coordinate axis vertically upwards, aligning the origin of coordinates with the Earth's surface, then to analyze free fall without an initial velocity, you can use the formula https://pandia.ru/text/78/086/images/image002_13.gif" width="151 "height="57 src=">

Near the surface of the Earth, in the absence of a noticeable influence of the atmosphere, the speed of a body thrown vertically upwards 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 a body at a certain height h can be found by the formula:

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The height of the body for some time, knowing the final speed

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IIIlevel. Solve problems. For 9 b. 9a solves from the problem book!

1. A ball is thrown vertically upwards with a speed of 18 m/s. What movement will he make in 3 seconds?

2. An arrow fired from a bow vertically upwards at a speed of 25 m/s hits the target after 2 s. What was the speed of the arrow when it hit the target?

3. A ball was fired vertically upwards from a spring pistol, which rose to a height of 4.9 m. With what speed did the ball fly out of the pistol?

4. The boy threw the ball vertically upwards and caught it after 2 s. What is the height of the ball and what is its initial speed?

5. With what initial speed should the body be thrown vertically upwards so that after 10 s it moves downwards 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 you need all the royal cavalry, all the royal army,

to Humpty, to Humpty, Humpty Dumpty,

Dumpty-Humpty collect "

(if it crashes only at 23 m/s?)

So is all the royal cavalry needed?

7. Now the thunder of sabers, spurs, sultan,
And the chamber junker caftan
Patterned - seductive beauties,
Was it not a temptation
When from the guard, others from the court
Came here on time!
Women shouted: hurrah!
And they threw caps into the air.

"Woe from Wit".

The girl Ekaterina threw her bonnet 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 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 falls 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 its maximum 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 ....