Updated:
Using several examples (which I initially solved, as usual, on otvet.mail.ru), consider a class of problems of elementary ballistics: the flight of a body launched at an angle to the horizon with a certain initial speed, without taking into account air resistance and the curvature of the earth's surface (that is, the direction We assume that the free fall acceleration vector g remains unchanged).
Task 1. The flight range of a body is equal to the height of its flight above the Earth's surface. At what angle is the body thrown? (for some reason some sources give the wrong answer - 63 degrees).
Let us denote the flight time as 2*t (then during t the body rises up, and during the next interval t it descends). Let the horizontal component of the velocity be V1, and the vertical component V2. Then flight range S = V1*2*t. Flight altitude H = g*t*t/2 = V2*t/2. We equate
S=H
V1*2*t = V2*t/2
V2/V1 = 4
The ratio of vertical and horizontal speeds is the tangent of the desired angle α, from which α = arctan(4) = 76 degrees.
Task 2. A body is thrown from the Earth's surface with a speed V0 at an angle α to the horizon. Find the radius of curvature of the body’s trajectory: a) at the beginning of the movement; b) at the top point of the trajectory.
In both cases, the source of curvilinear motion is gravity, that is, the acceleration of free fall g directed vertically downward. All that is required here is to find the projection g perpendicular to the current speed V, and equate it to the centripetal acceleration V^2/R, where R is the desired radius of curvature.
As can be seen from the figure, to start the movement we can write
gn = g*cos(a) = V0^2/R
whence the required radius R = V0^2/(g*cos(a))
For the top point of the trajectory (see figure) we have
g = (V0*cos(a))^2/R
whence R = (V0*cos(a))^2/g
Task 3. (variation on a theme) The projectile moved horizontally at a height h and exploded into two identical fragments, one of which fell to the ground at time t1 after the explosion. How long after the first fragment falls will the second fragment fall?
Whatever vertical velocity V the first fragment acquires, the second will acquire the same vertical velocity in magnitude, but directed in the opposite direction (this follows from the same mass of fragments and conservation of momentum). In addition, V is directed downwards, since otherwise the second fragment will fly to the ground BEFORE the first.
h = V*t1+g*t1^2/2
V = (h-g*t1^2/2)/t1
The second one will fly upward, lose vertical speed after time V/g, and then after the same time it will fly down to the initial height h, and its delay time t2 relative to the first fragment (not the flight time from the moment of explosion) will be
t2 = 2*(V/g) = 2h/(g*t1)-t1
updated 2018-06-03
Quote:
A stone is thrown at a speed of 10 m/s at an angle of 60° to the horizontal. Determine the tangential and normal acceleration of the body 1.0 s after the start of movement, the radius of curvature of the trajectory at this point in time, the duration and range of the flight. What angle does the total acceleration vector make with the velocity vector at t = 1.0 s
The initial horizontal speed Vg = V*cos(60°) = 10*0.5 = 5 m/s, and it does not change throughout the flight. Initial vertical velocity Vв = V*sin(60°) = 8.66 m/s. Flight time to the highest point t1 = Vв/g = 8.66/9.8 = 0.884 sec, which means the duration of the entire flight is 2*t1 = 1.767 sec. During this time, the body will fly horizontally Vg*2*t1 = 8.84 m (flight range).
After 1 second, the vertical speed will be 8.66 - 9.8*1 = -1.14 m/s (directed downward). This means the angle of speed to the horizon will be arctan(1.14/5) = 12.8° (down). Since the total acceleration here is the only and constant one (this is the acceleration of free fall g, directed vertically downwards), then the angle between the speed of the body and g at this point in time will be 90-12.8 = 77.2°.
Tangential acceleration is a projection g to the direction of the velocity vector, which means g*sin(12.8) = 2.2 m/s2. Normal acceleration is a projection perpendicular to the velocity vector g, it is equal to g*cos(12.8) = 9.56 m/s2. And since the latter is related to the speed and radius of curvature by the expression V^2/R, we have 9.56 = (5*5 + 1.14*1.14)/R, whence the desired radius R = 2.75 m.
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, the branches of which 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?
Here – initial speed of the body, – speed of the body at the moment of time t, s– horizontal flight range, h– the height above the surface of the earth from which a body is thrown horizontally with speed .
1.1.33. Kinematic equations for velocity projection:
1.1.34. Kinematic coordinate equations:
1.1.35. Body speed at a point in time t:
In the moment falling to the ground y = h, x = s(Fig. 1.9).
1.1.36. Maximum horizontal flight range:
1.1.37. Height above ground level, from which the body is thrown
horizontally:
Motion of a body thrown at an angle α to the horizontal
with initial speed
1.1.38. The trajectory is a parabola(Fig. 1.10). Curvilinear motion along a parabola is caused by the addition of two rectilinear motions: uniform motion along the horizontal axis and uniform motion along the vertical axis.
 |
| Rice. 1.10 |
( – initial speed of the body, – projections of velocity on the coordinate axes at the moment of time t, – body flight time, hmax– maximum body lifting height, smax– maximum horizontal flight range of the body).
1.1.39. Kinematic projection equations:
; 
1.1.40. Kinematic coordinate equations:
; 
1.1.41. Height of lifting the body to the top point of the trajectory:
At time , (Figure 1.11).
1.1.42. Maximum lifting height: 
1.1.43. Body flight time: 
At a moment in time , (Fig. 1.11).
1.1.44. Maximum horizontal body flight range:
1.2. Basic equations of classical dynamics
Dynamics(from Greek dynamis– force) is a branch of mechanics devoted to the study of the movement of material bodies under the influence of forces applied to them. Classical dynamics are based on Newton's laws . From these we obtain all the equations and theorems necessary for solving dynamics problems.
1.2.1. Inertial reporting system – This is a frame of reference in which the body is at rest or moves uniformly and rectilinearly.
1.2.2. Force- This is the result of the interaction of the body with the environment. One of the simplest definitions of force: the influence of a single body (or field) that causes acceleration. Currently, four types of forces or interactions are distinguished:
· gravitational(manifest in the form of universal gravitational forces);
· electromagnetic(existence of atoms, molecules and macrobodies);
· strong(responsible for the connection of particles in nuclei);
· weak(responsible for particle decay).
1.2.3. Principle of superposition of forces: if several forces act on a material point, then the resulting force can be found using the vector addition rule:
.
Body mass is a measure of body inertia. Any body exhibits resistance when trying to set it in motion or change the module or direction of its speed. This property is called inertia.
1.2.5. Pulse(momentum) is the product of mass T body by its speed v:
1.2.6. Newton's first law: Any material point (body) maintains a state of rest or uniform rectilinear motion until the influence of other bodies forces it (it) to change this state.
1.2.7. Newton's second law(basic equation of the dynamics of a material point): the rate of change of the momentum of the body is equal to the force acting on it (Fig. 1.11):
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| Rice. 1.11 | Rice. 1.12 |
The same equation in projections onto the tangent and normal to the trajectory of a point:
And 
.
1.2.8. Newton's third law: the forces with which two bodies act on each other are equal in magnitude and opposite in direction (Fig. 1.12):
1.2.9. Law of conservation of momentum for a closed system: the impulse of a closed system does not change over time (Fig. 1.13):
,
Where P– the number of material points (or bodies) included in the system.
 |  |
| Rice. 1.13 |
The law of conservation of momentum is not a consequence of Newton's laws, but is fundamental law of nature, which knows no exceptions, and is a consequence of the homogeneity of space.
1.2.10. The basic equation for the dynamics of translational motion of a system of bodies:
where is the acceleration of the center of inertia of the system; – total mass of the system from P material points.
1.2.11. Center of mass of the system material points (Fig. 1.14, 1.15):
.
Law of motion of the center of mass: the center of mass of a system moves like a material point, the mass of which is equal to the mass of the entire system and which is acted upon by a force equal to the vector sum of all forces acting on the system.
1.2.12. Impulse of a system of bodies:
where is the speed of the center of inertia of the system.
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| Rice. 1.14 | Rice. 1.15 |
1.2.13. Theorem on the motion of the center of mass: if the system is in an external stationary uniform field of forces, then no actions within the system can change the movement of the center of mass of the system:
.
1.3. Forces in mechanics
1.3.1. Body weight connection with gravity and ground reaction:
Acceleration of free fall (Fig. 1.16).
 |
| Rice. 1.16 |
Weightlessness is a state in which body weight is zero. In a gravitational field, weightlessness occurs when a body moves only under the influence of gravity. If a = g, That P = 0.
1.3.2. Relationship between weight, gravity and acceleration:
1.3.3. Sliding friction force(Fig. 1.17):
where is the sliding friction coefficient; N– normal pressure force.
1.3.5. Basic relations for a body on an inclined plane(Fig. 1.19). :
· friction force: 
;
· resultant force: ;
· rolling force: 
;
· acceleration:
 |
| Rice. 1.19 |
1.3.6. Hooke's law for a spring: spring extension X proportional to the elastic force or external force:
Where k– spring stiffness.
1.3.7. Potential energy of an elastic spring:
1.3.8. Work done by a spring:
1.3.9. Voltage– a measure of internal forces arising in a deformable body under the influence of external influences (Fig. 1.20):
where is the cross-sectional area of the rod, d– its diameter, – the initial length of the rod, – the increment in the length of the rod.
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| Rice. 1.20 | Rice. 1.21 |
1.3.10. Strain diagram – graph of normal stress σ = F/S from relative elongation ε = Δ l/l when the body is stretched (Fig. 1.21).
1.3.11. Young's modulus– quantity characterizing the elastic properties of the rod material:
1.3.12. Bar length increment proportional to voltage:
1.3.13. Relative longitudinal tension (compression):
1.3.14. Relative transverse tension (compression):
where is the initial transverse dimension of the rod.
1.3.15. Poisson's ratio– the ratio of the relative transverse tension of the rod to the relative longitudinal tension:
1.3.16. Hooke's law for a rod: the relative increment in the length of the rod is directly proportional to the stress and inversely proportional to Young’s modulus:
1.3.17. Volumetric potential energy density:
1.3.18. Relative shift ( fig1.22, 1.23 ):
where is the absolute shift.
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| Rice. 1.22 | Fig.1.23 |
1.3.19. Shear modulusG- a value that depends on the properties of the material and is equal to the tangential stress at which (if such huge elastic forces were possible).
1.3.20. Tangential elastic stress:
1.3.21. Hooke's law for shear:
1.3.22. Specific potential energy bodies in shear:
1.4. Non-inertial frames of reference
Non-inertial reference frame– an arbitrary reference system that is not inertial. Examples of non-inertial systems: a system moving in a straight line with constant acceleration, as well as a rotating system.
Inertial forces are caused not by the interaction of bodies, but by the properties of the non-inertial reference systems themselves. Newton's laws do not apply to inertial forces. Inertial forces are non-invariant with respect to the transition from one frame of reference to another.
In a non-inertial system, you can also use Newton's laws if you introduce inertial forces. They are fictitious. They are introduced specifically to take advantage of Newton's equations.
1.4.1. Newton's equation for a non-inertial reference frame
where is the acceleration of the body of mass T relative to a non-inertial system; – inertial force is a fictitious force due to the properties of the reference system.
1.4.2. Centripetal force– inertial force of the second kind, applied to a rotating body and directed radially to the center of rotation (Fig. 1.24):
,
where is the centripetal acceleration.
1.4.3. Centrifugal force– inertia force of the first kind, applied to the connection and directed radially from the center of rotation (Fig. 1.24, 1.25):
,
where is the centrifugal acceleration.
  |  |
| Rice. 1.24 | Rice. 1.25 |
1.4.4. Gravity acceleration dependence g depending on the latitude of the area is shown in Fig. 1.25.
Gravity is the result of the addition of two forces: and ; Thus, g(and therefore mg) depends on the latitude of the area:
,
where ω is the angular velocity of the Earth's rotation.
1.4.5. Coriolis force– one of the forces of inertia that exists in a non-inertial reference system due to rotation and the laws of inertia, manifesting itself when moving in a direction at an angle to the axis of rotation (Fig. 1.26, 1.27).
where is the angular velocity of rotation.
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| Rice. 1.26 | Rice. 1.27 |
1.4.6. Newton's equation for non-inertial reference systems taking into account all forces will take the form
where is the inertial force due to the translational motion of the non-inertial reference frame; And 
– two inertia forces caused by the rotational motion of the reference system; – acceleration of the body relative to a non-inertial reference frame.
1.5. Energy. Job. Power.
Conservation laws
1.5.1. Energy– a universal measure of various forms of movement and interaction of all types of matter.
1.5.2. Kinetic energy– function of the state of the system, determined only by the speed of its movement:
Kinetic energy of a body is a scalar physical quantity equal to half the product of mass m body per square of its speed.
1.5.3. Theorem on the change in kinetic energy. The work of the resultant forces applied to the body is equal to the change in the kinetic energy of the body, or, in other words, the change in the kinetic energy of the body is equal to the work A of all forces acting on the body.
1.5.4. Relationship between kinetic energy and momentum:
1.5.5. Work of force– quantitative characteristic of the process of energy exchange between interacting bodies. Mechanical work 
.
1.5.6. Constant force work:
If a body moves in a straight line and is acted upon by a constant force F, which makes a certain angle α with the direction of movement (Fig. 1.28), then the work of this force is determined by the formula:
,
Where F– force module, ∆r– module of displacement of the point of application of force, – angle between the direction of force and displacement.
If< /2, то работа силы положительна. Если >/2, then the work done by the force is negative. When = /2 (the force is directed perpendicular to the displacement), then the work done by the force is zero.
| Rice. 1.28 | Rice. 1.29 |
Constant force work F when moving along the axis x to a distance (Fig. 1.29) is equal to the projection of force on this axis multiplied by the displacement:
.
In Fig. Figure 1.27 shows the case when A < 0, т.к. >/2 – obtuse angle.
1.5.7. Elementary work d A strength F on elementary displacement d r is a scalar physical quantity equal to the scalar product of force and displacement:
1.5.8. Variable force work on trajectory section 1 – 2 (Fig. 1.30):
  |
| Rice. 1.30 |
1.5.9. Instantaneous power equal to the work done per unit time:
.
1.5.10. Average power for a period of time:
1.5.11. Potential energy body at a given point is a scalar physical quantity, equal to the work done by a potential force when moving a body from this point to another, taken as the zero potential energy reference.
Potential energy is determined up to some arbitrary constant. This is not reflected in the physical laws, since they include either the difference in potential energies in two positions of the body or the derivative of potential energy with respect to coordinates.
Therefore, the potential energy at a certain position is considered equal to zero, and the energy of the body is measured relative to this position (zero reference level).
1.5.12. Principle of minimum potential energy. Any closed system tends to transition to a state in which its potential energy is minimal.
1.5.13. The work of conservative forces equal to the change in potential energy
.
1.5.14. Vector circulation theorem: if the circulation of any force vector is zero, then this force is conservative.
The work of conservative forces along a closed contour L is zero(Fig. 1.31):
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| Rice. 1.31 |
1.5.15. Potential energy of gravitational interaction between the masses m And M(Fig. 1.32):
1.5.16. Potential energy of a compressed spring(Fig. 1.33):
  |   |
| Rice. 1.32 | Rice. 1.33 |
1.5.17. Total mechanical energy of the system equal to the sum of kinetic and potential energies:
E = E k + E P.
1.5.18. Body potential energy on high h above the ground
E n = mgh.
1.5.19. Relationship between potential energy and force:
Or 
or
1.5.20. Law of conservation of mechanical energy(for a closed system): the total mechanical energy of a conservative system of material points remains constant:
1.5.21. Law of conservation of momentum for a closed system of bodies:
1.5.22. Law of conservation of mechanical energy and momentum with an absolutely elastic central impact (Fig. 1.34):
Where m 1 and m 2 – body masses; and – the speed of the bodies before the impact.
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| Rice. 1.34 | Rice. 1.35 |
1.5.23. Speeds of bodies after an absolutely elastic impact (Fig. 1.35):
.
1.5.24. Speed of bodies after a completely inelastic central impact (Fig. 1.36):
1.5.25. Law of conservation of momentum when the rocket is moving (Fig. 1.37):
where and are the mass and speed of the rocket; and the mass and speed of the emitted gases.
 |  |
|
| Rice. 1.36 | Rice. 1.37 | |
1.5.26. Meshchersky equation for a rocket.
If the speed \(~\vec \upsilon_0\) is not directed vertically, then the movement of the body will be curvilinear.
Consider the motion of a body thrown horizontally from a height h with speed \(~\vec \upsilon_0\) (Fig. 1). We will neglect air resistance. To describe the movement, it is necessary to select two coordinate axes - Ox And Oy. The origin of the coordinates is compatible with the initial position of the body. From Figure 1 it is clear that υ 0x = υ 0 , υ 0y = 0, g x = 0, g y = g.
Then the motion of the body will be described by the equations:
\(~\upsilon_x = \upsilon_0,\ x = \upsilon_0 t; \qquad (1)\) \(~\upsilon_y = gt,\ y = \frac(gt^2)(2). \qquad (2) \)
Analysis of these formulas shows that in the horizontal direction the speed of the body remains unchanged, i.e. the body moves uniformly. In the vertical direction, the body moves uniformly with acceleration \(~\vec g\), i.e., the same as a body freely falling without an initial speed. Let's find the trajectory equation. To do this, from equation (1) we find the time \(~t = \frac(x)(\upsilon_0)\) and, substituting its value into formula (2), we obtain\[~y = \frac(g)(2 \ upsilon^2_0) x^2\] .
This is the equation of a parabola. Consequently, a body thrown horizontally moves along a parabola. The speed of the body at any moment of time is directed tangentially to the parabola (see Fig. 1). The velocity module can be calculated using the Pythagorean theorem:
\(~\upsilon = \sqrt(\upsilon^2_x + \upsilon^2_y) = \sqrt(\upsilon^2_0 + (gt)^2).\)
Knowing the altitude h with which the body is thrown, time can be found t 1 through which the body will fall to the ground. At this moment the coordinate y equal to height: y 1 = h. From equation (2) we find\[~h = \frac(gt^2_1)(2)\]. From here
\(~t_1 = \sqrt(\frac(2h)(g)). \qquad (3)\)
Formula (3) determines the flight time of the body. During this time the body will travel a distance in the horizontal direction l, which is called the flight range and which can be found based on formula (1), taking into account that l 1 = x. Therefore, \(~l = \upsilon_0 \sqrt(\frac(2h)(g))\) is the flight range of the body. The modulus of the body's velocity at this moment is \(~\upsilon_1 = \sqrt(\upsilon^2_0 + 2gh).\).
Literature
Aksenovich L. A. Physics in secondary school: Theory. Tasks. Tests: Textbook. allowance for institutions providing general education. environment, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsiya i vyhavanne, 2004. - P. 15-16.
Now it is not difficult for us to find out how the body will move if it is given an initial speed directed not at an arbitrary angle to the horizon, but horizontally. This is how, for example, a body moves when it comes off a horizontally flying plane (or is thrown from it).
We still believe that only gravity acts on such a body. She, as always, gives him a downward acceleration.
In the previous paragraph, we saw that a body thrown at an angle to the horizon, at a certain moment in time reaches the highest point of its trajectory (point B in Figure 134). At this moment, the speed of the body is directed horizontally.
We already know how the body moves after this. The trajectory of its movement is the right branch of the parabola shown in Figure 134. Any other body thrown horizontally will have a similar trajectory of movement. Figure 135 shows such a trajectory. It is also called a parabola, although it is only part of a parabola.
A body thrown horizontally moves along the branch of a parabola. Let's calculate the flight range for this body movement.
If a body is thrown from a height, we obtain from the formula the time during which it will fall
All the time while the body is falling down with acceleration, the vertical axis (Fig. 133) moves in the horizontal direction with speed
Therefore, during the fall it will move a distance
Hence,
This formula allows you to determine the flight range of a body thrown at a height horizontally with an initial speed
We looked at several examples of body motion under the influence of gravity. From them it is clear that in all cases the body moves with the acceleration imparted to it by the force of gravity. This acceleration is completely independent of whether the body is still moving in the horizontal direction or not. One can even say that in all these cases the body is in free fall.
Therefore, for example, a bullet fired by a shooter from a gun in a horizontal direction will fall to the ground at the same time as a bullet accidentally dropped by the shooter at the moment of the shot. But the dropped bullet will fall at the shooter’s feet, and the one flying out of the gun barrel will fall several hundred meters away from him.
The color insert shows a stroboscopic photograph of two balls, one of which falls vertically, and the second, simultaneously with the beginning of the fall of the first, is given speed in the horizontal direction. The photograph shows that at the same moments of time (moments of flashes of light) both balls are at the same height and, of course, reach the ground at the same time.
The trajectory of movement of bodies thrown horizontally or at an angle to the horizon can be clearly seen in a simple experiment. A bottle filled with water is placed at a certain height above the table and connected with a rubber tube to a tip equipped with a tap (Fig. 136). The jets released directly show the trajectories of water particles. By varying the angle at which the jet is released, you can ensure that the greatest range is achieved at an angle of 45°.
When considering the motion of a body thrown horizontally or at an angle to the horizon, we assumed that it was under the influence of gravity only. In reality this is not the case. Along with the force of gravity, the body is always affected by the force of resistance (friction) from the air. And it leads to a decrease in speed.
Therefore, the flight range of a body thrown horizontally or at an angle to the horizon is always less than what follows from the formulas,
received by us in this paragraph and § 55; the height of lift of a body thrown vertically is always less than that calculated by the formula given in § 21, etc.
The action of the resistance force also leads to the fact that the trajectory of a body thrown horizontally or at an angle to the horizon turns out to be not a parabola, but a more complex curve.
Exercise 33
Ignore friction when answering the questions in this exercise.
1. What is common in the motion of bodies thrown vertically, horizontally and at an angle to the horizon?
3. Is the acceleration of a body thrown horizontally the same at all points of its trajectory?
4. Is a body thrown horizontally in a state of weightlessness during its movement? What about a body thrown at an angle to the horizontal?
5. A body is thrown horizontally from a height of 2 m above the ground with a speed of 11 m/sec. How long will it take for it to fall? How far will the body travel in the horizontal direction?
6. A body is thrown with an initial speed of 20 m/sec in a horizontal direction at a height of 20 m above the Earth’s surface. At what distance from the point of throwing will it hit the ground? From what height must it be thrown at the same speed so that its flight range is doubled?
7. An airplane flies in a horizontal direction at an altitude of 10 km at a speed of 720 km/h. At what distance from the target (horizontally) must the pilot release the bomb to hit the target?