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:
| . | (3) |
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
When studying mechanical motion in physics, after becoming familiar with the uniform and uniformly accelerated movement of objects, they move on to considering the movement of a body at an angle to the horizon. In this article we will study this issue in more detail.
What is the movement of a body at an angle to the horizontal?
This type of object movement occurs when a person throws a stone into the air, a cannon fires a cannonball, or a goalkeeper kicks a soccer ball away from the goal. All such cases are considered by the science of ballistics.
The noted type of movement of objects in the air occurs along a parabolic trajectory. In the general case, carrying out the corresponding calculations is not a simple matter, since it is necessary to take into account air resistance, rotation of the body during flight, rotation of the Earth around its axis and some other factors.
In this article we will not take into account all these factors, but will consider the issue from a purely theoretical point of view. Nevertheless, the resulting formulas describe the trajectories of bodies moving over short distances quite well.
Obtaining formulas for the type of movement under consideration
Let's bring the bodies to the horizon at an angle. In this case, we will take into account only one single force acting on a flying object - gravity. Since it acts vertically downward (parallel to and against the y-axis), then, considering the horizontal and vertical components of the movement, we can say that the first will have the character of uniform rectilinear movement. And the second - uniformly slow (uniformly accelerated) rectilinear movement with acceleration g. That is, the velocity components through the value v 0 (initial speed) and θ (angle of direction of body motion) will be written as follows:
v x = v 0 *cos(θ)
v y = v 0 *sin(θ)-g*t
The first formula (for v x) is always valid. As for the second, one nuance should be noted here: the minus sign is placed in front of the product g*t only if the vertical component v 0 *sin(θ) is directed upward. In most cases, this is what happens, however, if you throw a body from a height, pointing it down, then in the expression for v y you should put a “+” sign in front of g*t.
Having integrated the formulas for the velocity components over time, and taking into account the initial height h of the body’s flight, we obtain equations for the coordinates:
x = v 0 *cos(θ)*t
y = h+v 0 *sin(θ)*t-g*t 2 /2
Flight range calculation
When considering in physics the movement of a body towards the horizon at an angle useful for practical application, it turns out to be the calculation of flight range. Let's define it.
Since this movement is a uniform movement without acceleration, it is enough to substitute the flight time into it and get the desired result. Flight range is determined solely by movement along the x-axis (parallel to the horizon).
The time a body remains in the air can be calculated by setting the y coordinate to zero. We have:
0 = h+v 0 *sin(θ)*t-g*t 2 /2
We solve this quadratic equation through the discriminant, we get:
D = b 2 - 4*a*c = v 0 2 *sin 2 (θ) - 4*(-g/2)*h = v 0 2 *sin 2 (θ) + 2*g*h,
t = (-b±√D)/(2*a) = (-v 0 *sin(θ)±√(v 0 2 *sin 2 (θ) + 2*g*h))/(-2* g/2) =
= (v 0 *sin(θ)+√(v 0 2 *sin 2 (θ) + 2*g*h))/g.
In the last expression, one root with a minus sign is discarded due to its insignificant physical significance. Substituting the flight time t into the expression for x, we obtain the flight range l:
l = x = v 0 *cos(θ)*(v 0 *sin(θ)+√(v 0 2 *sin 2 (θ) + 2*g*h))/g.
The easiest way to analyze this expression is if the initial height is zero (h=0), then we get a simple formula:
l = v 0 2 *sin(2*θ)/g
This expression indicates that the maximum flight range can be obtained if the body is thrown at an angle of 45 o (sin(2*45 o) = m1).
Maximum lifting height
In addition to the flight distance, it is also useful to find the height above the ground that the body can rise to. Since this type of movement is described by a parabola, the branches of which are directed downwards, the maximum lift height is its extremum. The latter is calculated by solving the equation for the t derivative of y:
dy/dt = d(h+v 0 *sin(θ)*t-g*t 2 /2)/dt = v 0 *sin(θ)-gt=0 =>
=> t = v 0 *sin(θ)/g.
Substituting this time into the equation for y, we get:
y = h+v 0 *sin(θ)*v 0 *sin(θ)/g-g*(v 0 *sin(θ)/g) 2 /2 = h + v 0 2 *sin 2 (θ)/( 2*g).
This expression indicates that the body will rise to its maximum height if it is thrown vertically upward (sin 2 (90 o) = 1).
In this article we will consider the analysis of a situation where a body is thrown at an angle to the horizontal. This could be throwing a stone by hand, firing a shell from a cannon, launching an arrow from a bow, and so on. All these situations are described in the same way from a mathematical point of view.
Feature of movement at an angle to the horizontal
What are the similarities between the above examples from a physics point of view? It lies in the nature of the forces acting on the body. During the free flight of a body, only two forces act on it:
- Gravity.
- Windage.
If the mass of the body is large enough and its shape is pointed (projectile, arrow), then air resistance can be neglected.
Thus, the movement of a body thrown at an angle to the horizon is a problem in which only gravity appears. It is this that determines the shape of the trajectory, which is described with good accuracy by a parabolic function.
Equations of motion along a parabolic trajectory. Speed
The body was thrown at an angle to the horizon. How can you describe his movement? Since the only force acting during the flight of a body is directed downward, its horizontal component is zero. This fact means that the horizontal movement of the object is uniquely determined by the initial conditions (throw or shot angle θ and speed v). The vertical movement of a body is a vivid example of uniformly accelerated motion, where the role of acceleration is played by the constant g (9.81 m/s2).
Taking into account the above, we can write two components for the speed of a flying body at time t:
v x = v * cos(θ);
v y = v * sin(θ) - g * t
As can be seen, the v x component does not depend on time and remains constant throughout the entire flight path (a consequence of the absence of external forces in the direction of the x axis). The component v y has a maximum at the initial moment of time. And then it begins to decrease until it becomes zero at the maximum point of takeoff of the body. After this, it changes sign and at the moment of falling it turns out to be equal to the modulus of the initial component v y, that is, v*sin(θ).
The written equations make it possible to determine the speed of a body thrown at an angle to the horizontal at any moment t. Its module will be equal to:
v = √ (v x 2 + v y 2) = √ (v 2 * cos 2 (θ) + v 2 * sin 2 (θ) - 2 * v* sin(θ) * g * t + g 2 * t 2) =
= √ (v 2 - 2 * v * sin(θ) * g * t + g 2 * t 2)
Equations of motion along a parabolic trajectory. Range of flight
The body was thrown at an angle to the horizon. How far will it fly? The range issue concerns the change in x coordinate. This value can be found by integrating both velocity components over time. As a result of integration we obtain the formulas:
x = v * cos(θ) * t + x 0 ;
y = v * sin(θ) * t - g * t 2 /2 + y 0
The difference between the coordinates x and x 0 is the flight range. If we assume that x 0 = 0, then the range will be equal to x, to find which you need to know how long t the body will be in the air.
The second equation allows you to calculate this time, provided that the value y 0 (the height h from which the body is thrown) is known. When the object completes its movement (falls to the ground), its y coordinate will become zero. Let's calculate the time when this will happen. We have:
v * sin(θ) * t - g * t 2 /2 + h = 0
Before us is a complete quadratic equality. We solve it through the discriminant:
D = v 2 * sin 2 (θ) - 4 * (-g/2) * h = v 2 * sin 2 (θ) + 2 * g * h;
t = (-v * sin(θ) ± √D)/(2 * (-g/2))
We discard the negative root. We get the following flight time:
t = (v * sin(θ) + √ (v 2 * sin 2 (θ) + 2 * g * h))/g
Now we substitute this value into the equation for flight range. We get:
x = v * cos(θ) * (v * sin(θ)+√ (v 2 * sin 2 (θ) + 2 * g * h))/g
If the body is thrown from the ground, that is, h = 0, then this formula will be significantly simplified. And it will look like:
x = 2 * v 2 * cos(θ) * sin(θ)/g = v 2 * sin(2 * θ)/g
The last expression was obtained using the relationship between the trigonometric functions of sine and cosine (reduction formula).
Since the sine has a maximum value for a right angle, then the maximum flight range is achieved when the body is thrown (shot) from the surface of the earth at an angle of 45°, and this range is equal to:
Height of a body thrown at an angle to the horizontal
Now let's determine another important parameter - the height to which a thrown object can rise. Obviously, for this it is enough to consider only the change in the y coordinate.
So, a body is thrown at an angle to the horizon, to what height will it fly up? This height will correspond to the equality of the velocity component v y to zero. We have the equation:
v y = v * sin(θ) - g * t = 0
Let's solve the equation. We get:
Now you need to substitute this time into the expression for the y coordinate. We get:
y = v * sin(θ) * t - g * t 2 /2 + h = v 2 * sin 2 (θ)/g - g/2* v 2 * sin 2 (θ)/g 2 + h =
V 2 * sin 2 (θ)/(2 * g) + h
This formula indicates that the maximum height, in contrast to the flight range, is obtained if the body is thrown strictly vertically (θ = 90). In this case we arrive at the formula:
It is interesting to note that in all the formulas given in this article, body weight does not appear. The characteristics of a parabolic trajectory do not depend on it, but only in the absence of air resistance.
The maximum range of a stone fired from a stationary catapult is S = 22.5 m. Find the maximum possible range of a stone fired from the same catapult mounted on a platform moving horizontally at a constant speed v = 15.0 m/s. Ignore air resistance, calculate free fall acceleration g = 10.0 m/s 2.
Solution: It is well known that the maximum flight range of a body thrown at an angle to the horizontal is achieved at an angle of departure equal to 45° and is determined by the formula:
Let us now consider the flight of a stone released from a moving catapult. Let us introduce a coordinate system whose axes are: X- directed horizontally, and Y— vertically. The origin of coordinates is compatible with the position of the catapult at the moment the stone is released.
To calculate the speed vector of the stone, it is necessary to take into account the horizontal speed of the catapult v = v o. Let's say that a catapult throws a stone at an angle α to the horizon. Then the components of the initial velocity of the stone in our coordinate system can be written as:
Substituting this expression into the first equation of system (3), we obtain the flight range of the stone:Secondly, it does not at all follow from (5) that S 1 will be maximum at α = 45°(this is true for (6), when v=0).
Proposing this problem for the Republican Olympiad, the authors were convinced that nine-tenths of the participants would receive formula (5) and then substitute the value into it α = 45°. However, to our regret, we were mistaken: not a single Olympian doubted that the maximum flight range is always (!) achieved at an angle of departure equal to 45°. This well-known fact has limited applicability: it is only true if:
a) do not take into account air resistance;
b) the take-off point and the fall point are at the same level;
c) the projectile is motionless.
Let's return to solving the problem. So we need to find the angle value α , at which S 1 determined by formula (5), is maximum. You can, of course, find the extremum of the function using the apparatus of differential calculus: find the derivative, set it equal to zero and, having solved the resulting equation, find the desired value α . However, given that the problem was proposed to 9th grade students, we will give its geometric solution. Let us take advantage of the fact that v = v o = 15 m/s.
Let's arrange the vectors v And v o as shown in fig. Since their lengths are equal, a circle with center at point O can be described around them. Then the length of the segment A.C. equal to v o + v o cos α(this is vxo), and the length of the segment B.C. equal to v o sin α(This vyo). Their product is equal to twice the area of the triangle ABC, or area of the triangle ABB 1.
Please note that it is the product that is included in the expression for flight range (5). In other words, the flight range is equal to the product of the area ΔАВВ 1 by a constant factor 2/g.
Now let’s ask ourselves: which of the triangles inscribed in a given circle has the maximum area? Naturally correct! Therefore, the desired value of the angle α = 60°.
Vector AB there is a vector of the total initial speed of the stone, it is directed at an angle 30° to the horizon (again, not at all 45°).
Thus, the final solution to the problem follows from formula (5), into which we should substitute α = 60°.
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 horizontal. 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 $S$: