The fall of a body thrown at an angle to the horizontal. Examples of solved problems in physics on the topic “free motion of a body thrown at an angle to the horizontal.” Maximum lifting height

Theory

If a body is thrown at an angle to the horizon, then in flight it is acted upon by the force of gravity and the force of air resistance. If the resistance force is neglected, then the only force left is gravity. Therefore, due to Newton's 2nd law, the body moves with acceleration equal to the acceleration of gravity; acceleration projections on the coordinate axes are equal a x = 0, and y= -g.

Any complex movement of a material point can be represented as a superposition of independent movements along the coordinate axes, and in the direction of different axes the type of movement may differ. In our case, the motion of a flying body can be represented as the superposition of two independent motions: uniform motion along the horizontal axis (X-axis) and uniformly accelerated motion along the vertical axis (Y-axis) (Fig. 1).

The body's velocity projections therefore change with time as follows:

,

where is the initial speed, α is the throwing angle.

The body coordinates therefore change like this:

With our choice of the origin of coordinates, the initial coordinates (Fig. 1) Then

The second time value at which the height is zero is zero, which corresponds to the moment of throwing, i.e. this value also has a physical meaning.

We obtain the flight range from the first formula (1). Flight range is the coordinate value X at the end of the flight, i.e. at a time equal to t 0. Substituting value (2) into the first formula (1), we get:

From this formula it can be seen that the greatest flight range is achieved at a throwing angle of 45 degrees.

The maximum lifting height of the thrown body can be obtained from the second formula (1). To do this, you need to substitute a time value equal to half the flight time (2) into this formula, because It is at the midpoint of the trajectory that the flight altitude is maximum. Carrying out calculations, we get

If a body is thrown at an angle to the horizon, then in flight it is acted upon by the force of gravity and the force of air resistance. If the resistance force is neglected, then the only force left is gravity. Therefore, due to Newton’s 2nd law, the body moves with acceleration equal to the acceleration of gravity; projections of acceleration onto the coordinate axes ax = 0, ay = - g.

Figure 1. Kinematic characteristics of a body thrown at an angle to the horizontal

Any complex movement of a material point can be represented as a superposition of independent movements along the coordinate axes, and in the direction of different axes the type of movement may differ. In our case, the motion of a flying body can be represented as the superposition of two independent motions: uniform motion along the horizontal axis (X-axis) and uniformly accelerated motion along the vertical axis (Y-axis) (Fig. 1).

The body's velocity projections therefore change with time as follows:

where $v_0$ is the initial speed, $(\mathbf \alpha )$ is the throwing angle.

With our choice of origin, the initial coordinates (Fig. 1) are $x_0=y_0=0$. Then we get:

(1)

Let's analyze formulas (1). Let us determine the time of movement of the thrown body. To do this, let's set the y coordinate equal to zero, because at the moment of landing the height of the body is zero. From here we get for the flight time:

The second time value at which the height is zero is zero, which corresponds to the moment of throwing, i.e. this value also has a physical meaning.

We obtain the flight range from the first formula (1). The flight range is the value of the x coordinate at the end of the flight, i.e. at time equal to $t_0$. Substituting value (2) into the first formula (1), we get:

From this formula it can be seen that the greatest flight range is achieved at a throwing angle of 45 degrees.

The maximum lifting height of the thrown body can be obtained from the second formula (1). To do this, you need to substitute a time value equal to half the flight time (2) into this formula, because It is at the midpoint of the trajectory that the flight altitude is maximum. Carrying out calculations, we get

From equations (1) one can obtain the equation of the body’s trajectory, i.e. an equation relating the x and y coordinates of a body during motion. To do this, you need to express time from the first equation (1):

and substitute it into the second equation. Then we get:

This equation is the motion trajectory equation. It can be seen that this is the equation of a parabola with its branches down, as indicated by the “-” sign in front of the quadratic term. It should be borne in mind that the throwing angle $\alpha $ and its functions are simply constants here, i.e. constant numbers.

A body is thrown with speed v0 at an angle $(\mathbf \alpha )$ to the horizon. Flight time $t = 2 s$. To what height Hmax will the body rise?

$$t_B = 2 s$$ $$H_max - ?$$

The law of body motion has the form:

$$\left\( \begin(array)(c) x=v_(0x)t \\ y=v_(0y)t-\frac(gt^2)(2) \end(array) \right.$ $

The initial velocity vector forms an angle $(\mathbf \alpha )$ with the OX axis. Hence,

\ \ \

A stone is thrown from the top of a mountain at an angle = 30$()^\circ$ to the horizon with an initial speed of $v_0 = 6 m/s$. Inclined plane angle = 30$()^\circ$. At what distance from the point of throwing will the stone fall?

$$ \alpha =30()^\circ$$ $$v_0=6\ m/s$$ $$S - ?$$

Let's place the origin of coordinates at the throwing point, OX - along the inclined plane downwards, OY - perpendicular to the inclined plane upwards. Kinematic characteristics of movement:

Law of motion:

$$\left\( \begin(array)(c) x=v_0t(cos 2\alpha +g\frac(t^2)(2)(sin \alpha \ )\ ) \\ y=v_0t(sin 2 \alpha \ )-\frac(gt^2)(2)(cos \alpha \ ) \end(array) \right.$$ \

Substituting the resulting value $t_В$, we find $S$:

If air resistance can be neglected, then a body thrown in any way moves with the acceleration of gravity.

Let us first consider the motion of a body thrown horizontally with a speed v_vec0 from a height h above the earth's surface (Fig. 11.1).

In vector form, the dependence of the speed of a body on time t is expressed by the formula

In projections on the coordinate axes:

v x = v 0 , (2)
v y = –gt. (3)

1. Explain how formulas are obtained from (2) and (3)

x = v 0 t, (4)
y = h – gt 2 /2. (5)

We see that the body seems to be performing two types of motion simultaneously: it moves uniformly along the x axis, and uniformly accelerated along the y axis without an initial speed.

Figure 11.2 shows the position of the body at regular intervals. Below is shown the position at the same instants of time of a body moving rectilinearly uniformly with the same initial speed, and on the left is the position of a freely falling body.

We see that a body thrown horizontally is always on the same vertical with a uniformly moving body and on the same horizontal with a freely falling body.

2. Explain how from formulas (4) and (5) we obtain expressions for time tfloor and body flight distance l:

Clue. Take advantage of the fact that at the moment of falling y = 0.

3. A body is thrown horizontally from a certain height. In which case will the body's flight range be greater: when the initial speed increases by 4 times or when the initial height increases by the same number? How many times more?

Movement trajectories

In Figure 11.2, the trajectory of a body thrown horizontally is depicted by a red dashed line. It resembles a branch of a parabola. Let's check this assumption.

4. Prove that for a body thrown horizontally, the equation of the trajectory of motion, that is, the dependence y(x), is expressed by the formula

Clue. Using formula (4), express t in terms of x and substitute the found expression into formula (5).

Formula (8) is indeed a parabolic equation. Its vertex coincides with the initial position of the body, that is, it has coordinates x = 0; y = h, and the branch of the parabola is directed downward (this is indicated by the negative coefficient in front of x 2).

5. The dependence y(x) is expressed in SI units by the formula y = 45 – 0.05x 2.
a) What are the initial height and initial velocity of the body?
b) What are the flight time and distance?

6. A body is thrown horizontally from a height of 20 m with an initial speed of 5 m/s.
a) How long will the body’s flight last?
b) What is the flight range?
c) What is the speed of the body just before it hits the ground?
d) At what angle to the horizon will the body’s velocity be directed immediately before hitting the ground?
e) What formula expresses in SI units the dependence of the velocity modulus of a body on time?

2. Movement of a body thrown at an angle to the horizontal

Figure 11.3 schematically shows the initial position of the body, its initial speed 0 (at t = 0) and acceleration (gravitational acceleration).

Initial velocity projections

v 0x = v 0 cos α, (9)
v 0y = v 0 sin α. (10)

To shorten subsequent entries and clarify their physical meaning, it is convenient to retain the notation v 0x and v 0y before obtaining the final formulas.

The velocity of the body in vector form at time t is also in this case expressed by the formula

However, now in projections on the coordinate axes

v x = v 0x , (11)
vy = v 0y – gt. (12)

7. Explain how the following equations are obtained:

x = v 0x t, (13)
y = v 0y t – gt 2 /2. (14)

We see that in this case, too, the thrown body seems to be involved in two types of motion simultaneously: it moves uniformly along the x axis, and uniformly accelerates along the y axis with an initial speed, like a body thrown vertically upward.

Trajectory of movement

Figure 11.4 schematically shows the position of a body thrown at an angle to the horizontal at regular intervals. Vertical lines emphasize that the body moves uniformly along the x-axis: adjacent lines are at equal distances from each other.

8. Explain how to obtain the following equation for the trajectory of a body thrown at an angle to the horizontal:

Formula (15) is the equation of a parabola whose branches are directed downward.

The trajectory equation can tell us a lot about the motion of a thrown body!

9. The dependence y(x) is expressed in SI units by the formula y = √3 * x – 1.25x 2.
a) What is the horizontal projection of the initial velocity?
b) What is the vertical projection of the initial velocity?
c) At what angle is the body thrown to the horizon?
d) What is the initial speed of the body?

The parabolic shape of the trajectory of a body thrown at an angle to the horizon is clearly demonstrated by a stream of water (Fig. 11.5).

Ascent time and entire flight time

10. Using formulas (12) and (14), show that the body’s rise time t under and the entire flight time t floor are expressed by the formulas

Clue. At the top point of the trajectory v y = 0, and at the moment the body falls its coordinate is y = 0.

We see that in this case (the same as for a body thrown vertically upward) the entire flight time t floor is 2 times longer than the rise time t under. And in this case, when viewing the video in reverse, the rise of the body will look exactly like its descent, and the descent will look exactly like its rise.

Altitude and flight range

11. Prove that the lift height h and flight range l are expressed by the formulas

Clue. To derive formula (18), use formulas (14) and (16) or formula (10) from § 6. Displacement during rectilinear uniformly accelerated motion; to derive formula (19), use formulas (13) and (17).

Please note: the lifting time of the body tunder, the entire flight time tfloor and the lifting height h depend only on the vertical projection of the initial speed.

12. To what height did the soccer ball rise after being hit if it fell to the ground 4 s after the hit?

13. Prove that

Clue. Use formulas (9), (10), (18), (19).

14. Explain why, at the same initial speed v 0, the flight range l will be the same at two angles α 1 and α 2, related by the relation α 1 + α 2 = 90º (Fig. 11.6).

Clue. Use the first equality in formula (21) and the fact that sin α = cos(90º – α).

15. Two bodies thrown at the same time and with the same initial value and one point. The angle between the initial velocities is 20º. At what angles to the horizon were the bodies thrown?

Maximum flight range and altitude

At the same absolute initial speed, the flight range and altitude are determined only by the angle α. How to choose this angle so that the flight range or altitude is maximum?

16. Explain why the maximum flight range is achieved at α = 45º and is expressed by the formula

l max = v 0 2 /g. (22)

17.Prove that the maximum flight altitude is expressed by the formula

h max = v 0 2 /(2g) (23)

18. A body thrown at an angle of 15º to the horizontal fell at a distance of 5 m from the starting point.
a) What is the initial speed of the body?
b) To what height did the body rise?
c) What is the maximum flight range at the same absolute initial speed?
d) To what maximum height could this body rise at the same absolute initial speed?

Dependence of speed on time

When ascending, the speed of a body thrown at an angle to the horizontal decreases in absolute value, and when descending, it increases.

19. A body is thrown at an angle of 30º to the horizontal with an initial speed of 10 m/s.
a) How is the dependence vy(t) expressed in SI units?
b) How is the dependence v(t) expressed in SI units?
c) What is the minimum speed of a body during flight?
Clue. Use formulas (13) and (14), as well as the Pythagorean theorem.

Additional questions and tasks

20. Throwing pebbles at different angles, Sasha discovered that he could not throw the pebble further than 40 m. What is the maximum height that Sasha can throw the pebble?

21. There was a pebble stuck between the rear dual tires of a truck. At what distance from the truck should the car following it be driven so that this pebble, if it falls off, does not cause harm to it? Both cars are traveling at a speed of 90 km/h.
Clue. Go to the frame of reference associated with any of the cars.

22. At what angle to the horizon should a body be thrown in order to:
a) was the flight altitude equal to the range?
b) the flight altitude was 3 times greater than the range?
c) the flight range was 4 times greater than the altitude?

23. A body is thrown with an initial speed of 20 m/s at an angle of 60º to the horizontal. At what time intervals after the throw will the body's speed be directed at an angle of 45º to the horizontal?

Let us consider, as an example of the application of the derived formulas, the movement of a body thrown at an angle to the horizon in the absence of air resistance. Let's say, on a mountain, at a height above sea level, there is a cannon guarding the coastal waters. Let the projectile be fired at an angle to the horizon with an initial speed from a point, the position of which is determined by the radius vector (Fig. 2.16).

Rice. 2.16. Movement of a body thrown at an angle to the horizontal

Addition.

Derivation of the equations of motion of a material point in gravity field

Let us write the equation of motion (equation Newton's second law):

this means that bodies - material points - of any mass under the same initial conditions will move in a uniform gravitational field in the same way. Let's project equation (2.7.2) on the axis of the Cartesian coordinate system. Horizontal axis OH shown in Fig. 13 dotted line, axis OY let's draw through the point ABOUT vertically upward, and the horizontal axis OZ, also passing through the point ABOUT, direct it perpendicular to the vector towards us. We get:

The vertical direction, by definition, is the direction of the vector, therefore its projections onto the horizontal axes OX And OY are equal to zero. The second equation takes into account that the vector is directed downward and the axis OY- up.

Rice. 2.17. The motion of a body thrown at an angle to the horizontal.

Let's add initial conditions to the equations of motion, which determine the position and speed of the body at the initial moment of time t 0, let t0 = 0. Then, according to Fig. 2.7.4

If the derivative of some function is equal to zero, then the function is constant, respectively, from the first and third equations (2.7.3) we obtain:

In the second equation (2.7.3) the derivative is equal to a constant, which means that the function depends linearly on its argument, that is

Combining (2.7.7) and (2.7.9), we obtain the final expressions for the dependences of velocity projections on the coordinate axes on time:

The third equation (2.7.11) shows that the trajectory of the body is flat and lies entirely in the plane XOY, is the vertical plane defined by the vectors and . Obviously, the last statement is general: no matter how the directions of the coordinate axes are chosen, the trajectory of a body thrown at an angle to the horizon is flat, it always lies in the plane determined by the initial velocity vector and the free fall acceleration vector.

If the three equations (2.7.10) are multiplied by the unit vectors of the axes , , and and added, and then the same is done with the three equations (2.7.11), then we obtain the time dependence of the particle velocity vector and its radius vector. Taking into account the initial conditions we have:

Formulas (2.7.12) and (2.7.13) could be obtained immediately, directly from (2.7.2), if we take into account that the acceleration of gravity is a constant vector. If the acceleration - the derivative of the velocity vector - is constant, then the velocity vector depends linearly on time, and the radius vector, the time derivative of which is the velocity vector linearly dependent on time, depends quadratically on time. This is written in relations (2.7.12) and (2.7.13) with constants - constant vectors - selected according to the initial conditions in the form (2.7.4).

From (2.7.13), in particular, it is clear that the radius vector is the sum of three vectors that add up according to the usual rules, which is clearly shown in Fig. 2.18.

Rice. 2.18. Representation of the radius vector r(t) at an arbitrary time t as a sum of three vectors

These vectors are:

Here the principle of independence of motions, known in other areas of physics as superposition principle(overlays). Generally speaking, according to the principle of superposition, the resulting effect of several influences is the sum of the effects of each influence separately. It is a consequence of the linearity of the equations of motion.

Video 2.3. Independence of horizontal and vertical movements when moving in a field of gravity.

Let's place the origin at the throwing point. Now =0 , the axes, as before, will be rotated so that the axis 0x was horizontal, the axis 0у- vertical, and the initial speed lay in the plane x0y(Fig. 2.19).

Rice. 2.19. Projections of initial velocity onto coordinate axes

Let's project onto the coordinate axes (see (2.7.11)):

Flight path. If we exclude time from the system of obtained equations t, then we obtain the trajectory equation:

This is the equation of a parabola whose branches are directed downward.

Flight range when firing from a height h . At the moment the body falls (the projectile hits a target located on the surface of the sea). The horizontal distance from the gun to the target is equal to . Substituting ; into the trajectory equation, we obtain a quadratic equation for flight range:

The quadratic equation has two solutions (in this case, positive and negative). We need a positive solution. The standard expression for the root of the quadratic equation of our problem can be reduced to the form:

is achieved at , if h = 0.

Maximum flight range. When shooting from a mountain high, this is no longer the case. Let's find the angle at which the maximum flight range is achieved. The dependence of flight range on angle is quite complex, and instead of differentiation to find the maximum, we will proceed as follows. Let's imagine that we increase the starting angle. First, the flight range increases (see formula (2.7.15)), reaches a maximum value and begins to fall again (to zero when shooting vertically upward). Thus, for each flight range, except for the maximum, there are two directions of initial speed.

Let us turn again to the quadratic equation of relativity of flight range and consider it as an equation for the angle. Considering that

let's rewrite it in the form:

We have again obtained a quadratic equation, this time for an unknown quantity. The equation has two roots, which corresponds to two angles at which the flight range is equal to . But when , both roots must coincide. This means that the discriminant of the quadratic equation is equal to zero:

where does the result follow?

When this result reproduces formula (2.7.16)

Usually the altitude is much less than the flight range on the plain. When the square root can be approximated by the first terms of the Taylor series expansion and we obtain the approximate expression

that is, the firing range increases approximately by the height of the gun's elevation.

When l = lmax, And a = a max , as already noted, the discriminant of the quadratic equation is equal to zero, respectively, its solution has the form:

Since the tangent is less than one, the angle at which maximum flight range is achieved is less.

Maximum lift height above the starting point. This value can be determined from the equality to zero of the vertical component of velocity at the top point of the trajectory

In this case, the horizontal component of the velocity is not equal to zero, therefore