Consider, as an example of the application of the derived formulas, the motion of a body thrown at an angle to the horizon in the absence of air resistance. 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 velocity from a point whose position is determined by the radius vector (Fig. 2.16).
Rice. 2.16. Movement of a body thrown at an angle to the horizon
Addition.
Derivation of the equations of motion of a material point in the field of gravity
Let's write the equation of motion (the equation of 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 us project the equation (2.7.2) on the axes of the Cartesian coordinate system. Horizontal axis OH shown in fig. 13 dashed axis OY pass through the point O vertically upwards, and the horizontal axis oz also passing through the point O, direct perpendicular to the vector at us. We get:
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The vertical direction, by definition, is the direction of the vector, so its projections on the horizontal axes OX and OY are equal to zero. The second equation takes into account that the vector is directed downwards, and the axis OY- up.
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Rice. 2.17. The motion of a body thrown at an angle to the horizon.
Let us add to the equations of motion the initial conditions that determine the position and velocity of the body at the initial moment of time t0, let be 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 the velocity projections on the coordinate axes on time:
The third equation (2.7.11) shows that the trajectory of the body is flat, 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 gravitational acceleration vector.
If three equations (2.7.10) are multiplied by the unit vectors of the axes , , and and added, and then the same is done with three equations (2.7.11), then we get the time dependence of the particle velocity vector and its radius vector. Taking into account the initial conditions, we have:
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Formulas (2.7.12) and (2.7.13) could be obtained immediately, directly from (2.7.2), given that the gravitational acceleration 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 that depends linearly on time, depends quadratically on time. This is written in relations (2.7.12) and (2.7.13) with constants - constant vectors - chosen according to the initial conditions in the form (2.7.4).
From (2.7.13), in particular, it can be seen that the radius vector is the sum of three vectors added 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 the sum of three vectors
These vectors are:
Here the principle of independence of motion, known in other areas of physics as superposition principle(overlays). Generally speaking, according to the principle of superposition, the net effect of several actions is the sum of the effects of each action taken 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 the field of gravity.
Let's place the origin at the drop point. Now =0 , the axes, as before, will be rotated so that the axis 0x was horizontal, the axis 0y- vertical, and the initial speed lay in the plane x0y(Fig. 2.19).
Rice. 2.19. Projections of the initial velocity on the coordinate axes
We project onto the coordinate axes (see (2.7.11)):
Flight path. If time is excluded from the system of equations obtained t, then we get the trajectory equation:
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This is the equation of a parabola, the branches of which are directed downwards.
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 the flight range:
A quadratic equation has two solutions (in this case, positive and negative). We need a positive decision. 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 shot from a high mountain, this is no longer the case. Find the angle at which the maximum flight range is reached. The dependence of the flight range on the angle is quite complicated, and instead of differentiating to find the maximum, we will do the following. Let's imagine that we increase the initial angle . First, the flight range increases (see formula (2.7.15)), reaches its maximum value and starts to fall again (to zero when fired vertically upwards). Thus, for each flight range, except for the maximum, there are two directions of the initial speed.
Let us turn again to the quadratic equation for the flight distance relativity and consider it as an equation for the angle . Given that
let's rewrite it in the form:
We again got 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 . But when , both roots must match. This means that the discriminant of the quadratic equation is equal to zero:
where does the result come from
With this result reproduces the formula (2.7.16)
Usually the height is much less than the flight range on the plain. For , the square root can be approximated by the first terms of the Taylor series expansion, and we obtain the approximate expression
that is, the range of the shot increases by approximately the height of the gun.
When l = l max , 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 the maximum flight range is reached is smaller.
The maximum climb height above the starting point. This value can be determined from the equality to zero of the vertical velocity component at the top of the trajectory
In this case, the horizontal component of the velocity is not equal to zero, therefore
There were 3 seconds left until the end of the final match of the basketball tournament of the 1972 Munich Olympics. The Americans - the US team - were already celebrating the victory! Our team - the USSR national team - won about 10 points against the great dream Team...
A few minutes before the end of the match. But, having lost all the advantage in the end, she was already losing one point 49:50. What happened next was incredible! Ivan Edeshko throws the ball from behind the end line across the entire area under the Americans' ring, where our center Alexander Belov receives the ball surrounded by two opponents and puts it into the basket. 51:50 - we are Olympic champions!!!
I, being a child then, experienced the strongest emotions - first disappointment and resentment, then crazy delight! The emotional memory of this episode is etched into my mind for the rest of my life! Watch the video on the Internet for the request "Alexander Belov's golden throw", you will not regret it.
The Americans then did not admit defeat and refused to receive silver medals. Is it possible to do in three seconds what our players did? Let's remember physics!
In this article, we will consider the motion of a body thrown at an angle to the horizon, create an Excel program for solving this problem with various combinations of input data, and try to answer the above question.
This is a fairly well-known problem in physics. In our case, the body thrown at an angle to the horizon is a basketball. We will calculate the initial speed, time and trajectory of the ball thrown across the entire court by Ivan Edeshko and falling into the hands of Alexander Belov.
Mathematics and physics of basketball flight.
The formulas below and the calculation inexcel are universal for a wide range of problems about bodies thrown at an angle to the horizon and flying along a parabolic trajectory without taking into account the effect of air friction.
The calculation scheme is shown in the figure below. Launch MS Excel or OOo Calc.
Initial data:
1. Since we are on planet Earth and are considering a ballistic problem - the movement of bodies in the Earth's gravity field, then first of all we write down the main characteristic of the gravitational field - the free fall acceleration g in m/s 2
to cell D3: 9,81
2. The size of the basketball court is 28 meters long and 15 meters wide. The flight distance of the ball almost across the court to the ring from the opposite end line horizontally x write in meters
to cell D4: 27,000
3. If we assume that Edeshko made the throw from a height of about two meters, and Belov caught the ball just somewhere at the level of the ring, then with a basketball hoop height of 3.05 meters, the distance between the points of departure and arrival of the ball will be vertically 1 meter. Let's write down the vertical displacement y in meters
to cell D5: 1,000
4. According to my measurements on the video, the angle of departure of the ball α 0 from the hands of Edeshko did not exceed 20 °. Enter this value
to cell D6: 20,000
Calculation results:
Basic equations describing the motion of a body thrown at an angle to the horizon without taking into account air resistance:
x =v0* cos α 0 *t
y =v0*sin α 0 *t -g *t 2 /2
5. Let's express the time t from the first equation, substitute into the second and calculate the initial speed of the ball v 0 in m/s
in cell D8: =(D3*D4^2/2/COS (RADIANS(D6))^2/(D4*TAN (RADIANS(D6))-D5))^0.5 =21,418
v0 =(g *x 2 /(2*(cosα 0 ) 2 *(x *tgα 0 -y)) 0.5
6. Time of flight of the ball from the hands of Edeshko to the hands of Belov t calculate in seconds, knowing now v 0 , from the first equation
in cell D9: =D4/D8/COS (RADIANS(D6)) =1,342
t = x /(v 0 * cosα 0 )
7. Find the angle of direction of the ball's speed α i at the point of interest to us. To do this, we write the initial pair of equations in the following form:
y =x *tgα 0 -g *x 2 /(2*v 0 2*(cosα 0 ) 2)
This is the equation of a parabola - the flight path.
We need to find the angle of inclination of the tangent to the parabola at the point of interest to us - this will be the angle α i. To do this, take the derivative, which is the tangent of the slope of the tangent:
y' =tgα 0 -g *x /(v 0 2*(cosα 0 ) 2)
Calculate the angle of arrival of the ball in the hands of Belov α i in degrees
in cell D10: =ATAN (TAN (RADIANS(D6)) -D3*D4/D8^2/COS (RADIANS(D6))^2)/PI()*180 =-16,167
α i = arctgy ’ = arctg(tgα 0 — g * x /(v 0 2 *(cosα 0 ) 2))
The calculation in excel, in principle, is completed.
Other payment options:
Using the written program, you can quickly and easily perform calculations with other combinations of initial data.
Let, given a horizontal x = 27 meters , vertical y = 1 meter flight range and initial speed v 0 = 25 m/s.
It is required to find the flight time t and departure angles α 0 and arrival α i
Let's use the service MS Excel "Selection of the parameter". I have repeatedly described in detail in several blog articles how to use it. You can read more about using this service.
We set the value in cell D8 to 25,000 by changing the selection of the value in cell D6. The result is in the picture below.
The initial data in this version of the calculation in excel (as, indeed, in the previous one) are highlighted in blue frames, and the results are circled in red rectangular frames!
Setting the tableexcel some value of interest in one of the cells with a light yellow fill, by selecting a changed value in one of the cells with a light turquoise fill, in the general case, you can get ten different options for solving the problem of the motion of a body thrown at an angle to the horizon with ten different sets source data!!!
Answer to the question:
Let's answer the question posed at the beginning of the article. The ball sent by Ivan Edeshko flew to Belov, according to our calculations, in 1.342 seconds. Alexander Belov caught the ball, landed, jumped up and threw it. For all this he had a "sea" of time - 1.658s! This is really enough time with a margin! A detailed view of the video frame by frame confirms the above. Three seconds were enough for our players to deliver the ball from their front line to the opponents' backboard and throw it into the ring, writing their names in basketball history with gold!
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Kinematics is easy!
After the throw, in flight, gravity acts on the body Ft and the force of air resistance Fс.
If the movement of the body occurs at low speeds, then the air resistance force is usually not taken into account when calculating.
So, we can assume that only gravity acts on the body, which means that the motion of the thrown body is free fall.
If this is free fall, then the acceleration of the thrown body is equal to the acceleration of free fall g.
At low altitudes relative to the Earth's surface, the force of gravity Ft practically does not change, so the body moves with constant acceleration.
So, the motion of a body thrown at an angle to the horizon is a variant of free fall, i.e. movement with constant acceleration and curvilinear trajectory(since the velocity and acceleration vectors do not coincide in direction).
Formulas of this movement in vector form: the body's trajectory is a parabola lying in a plane passing through the vectors Fт and Vo .
The point of origin of the thrown body is usually chosen as the origin of coordinates.
At any moment in time, the change in the speed of the body in the direction coincides with the acceleration.
The velocity vector of the body at any point of the trajectory can be decomposed into 2 components: the vector V x and the vector V y .
At any moment in time, the speed of the body will be determined as the geometric sum of these vectors:
According to the figure, the projections of the velocity vector on the coordinate axes OX and OY look like this:
Calculation of body speed at any moment of time:
Calculation of the displacement of the body at any time:
Each point of the body motion trajectory corresponds to X and Y coordinates:
Calculation formulas for the coordinates of the thrown body at any time:
From the motion equation, formulas can be derived for calculating the maximum flight range L:
and maximum flight altitude H:
P.S.
1. With equal initial speeds Vo, the flight range:
- increases if the initial throwing angle is increased from 0 o to 45 o ,
- Decreases if the initial throwing angle is increased from 45 o to 90 o .
2. With equal initial throwing angles, the flight range L increases with an increase in the initial speed Vo. 
3. A special case of the motion of a body thrown at an angle to the horizon is motion of a body thrown horizontally, while the initial throwing angle is zero.
If a body is thrown at an angle to the horizon, then in flight it is affected by gravity and air resistance. If the resistance force is neglected, then the only force left is the force of gravity. Therefore, due to Newton's 2nd law, the body moves with an acceleration equal to the acceleration of free fall; acceleration projections on the coordinate axes ax = 0, ay = - g.
Figure 1. Kinematic characteristics of a body thrown at an angle to the horizon
Any complex movement of a material point can be represented as an imposition 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 a 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 velocity projections of the body 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 us analyze formulas (1). Let us determine the time of motion of the thrown body. To do this, we 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 value of the time at which the height is equal to zero is equal to zero, which corresponds to the moment of throwing, i.e. this value also has a physical meaning.
The flight range is obtained from the first formula (1). Flight range is the value of the x-coordinate at the end of the flight, i.e. at the moment of time equal to $t_0$. Substituting the value (2) into the first formula (1), we obtain:
From this formula it can be seen that the greatest flight range is achieved at a throw angle of 45 degrees.
The highest lifting height of the thrown body can be obtained from the second formula (1). To do this, you need to substitute in this formula the value of time equal to half the flight time (2), 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 trajectory, i.e. an equation relating the x and y coordinates of a body during motion. To do this, you need to express the time from the first equation (1):
and substitute it into the second equation. Then we get:
This equation is the trajectory equation. It can be seen that this is the equation of a parabola with branches down, as indicated by the “-” sign in front of the quadratic term. It should be kept in mind that the throwing angle $\alpha $ and its functions are just constants here, i.e. constant numbers.
A body is thrown with velocity 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 is:
$$\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 throw will the stone fall?
$$ \alpha =30()^\circ$$ $$v_0=6\ m/s$$ $$S - ?$$
Let's place the origin of coordinates at the drop point, OX - along the inclined plane down, OY - perpendicular to the inclined plane up. 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 of $t_B$, we find $S$:
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