A body thrown at an angle to the horizontal theory. Free fall of bodies

What is free fall? This is the fall of bodies to the Earth in the absence of air resistance. In other words, falling into the void. Of course, the absence of air resistance is a vacuum, which cannot be found on Earth in normal conditions. Therefore, we will not take the force of air resistance into account, considering it so small that it can be neglected.

Acceleration of gravity

Carrying out his famous experiments on the Leaning Tower of Pisa, Galileo Galilei found out that all bodies, regardless of their mass, fall to the Earth in the same way. That is, for all bodies the acceleration free fall the same. According to legend, the scientist then dropped balls of different masses from the tower.

Acceleration of gravity

Gravity acceleration is the acceleration with which all bodies fall to the Earth.

The acceleration of gravity is approximately 9.81 m s 2 and is denoted by the letter g. Sometimes, when accuracy is not fundamentally important, the acceleration of gravity is rounded to 10 m s 2.

The Earth is not a perfect sphere, and at different points on the earth's surface, depending on the coordinates and altitude above sea level, the value of g varies. Thus, the greatest acceleration of gravity is at the poles (≈ 9.83 m s 2), and the smallest is at the equator (≈ 9.78 m s 2).

Free fall body

Let's look at a simple example of free fall. Let some body fall from a height h with zero initial speed. Let's say we raised the piano to a height h and calmly released it.

Free fall is a rectilinear movement with constant acceleration. Let's direct the coordinate axis from the point of initial position of the body to the Earth. Using kinematics formulas for rectilinear uniformly accelerated motion, we can write:

h = v 0 + g t 2 2 .

Since the initial speed is zero, we rewrite:

From here we find the expression for the time of falling of a body from a height h:

Taking into account that v = g t, we find the speed of the body at the moment of falling, that is, the maximum speed:

v = 2 h g · g = 2 h g .

Similarly, we can consider the motion of a body thrown vertically upward with a certain initial speed. For example, we throw a ball up.

Let the coordinate axis be directed vertically upward from the point of throwing the body. This time the body moves equally slow, losing speed. At the highest point the speed of the body is zero. Using kinematics formulas, we can write:

Substituting v = 0, we find the time for the body to rise to its maximum height:

The time of fall coincides with the time of rise, and the body will return to Earth after t = 2 v 0 g.

Maximum lifting height of a body thrown vertically:

Let's take a look at the picture below. It shows graphs of body velocities for three cases of motion with acceleration a = - g. Let's consider each of them, having previously clarified that in this example all numbers are rounded, and the acceleration of free fall is assumed to be 10 m s 2.

The first graph is a body falling from a certain height without initial speed. Fall time tp = 1 s. From the formulas and from the graph it is easy to see that the height from which the body fell is h = 5 m.

The second graph is the movement of a body thrown vertically upward with an initial speed v 0 = 10 m s. Maximum lifting height h = 5 m. Rising time and falling time t p = 1 s.

The third graph is a continuation of the first. The falling body bounces off the surface and its speed sharply changes sign to the opposite. Further movement of the body can be considered according to the second graph.

The problem of the free fall of a body is closely related to the problem of the motion of a body thrown at a certain angle to the horizon. Thus, movement along a parabolic trajectory can be represented as the sum of two independent movements relative to the vertical and horizontal axes.

Along the O Y axis the body moves uniformly with acceleration g, the initial speed of this movement is v 0 y. The movement along the O X axis is uniform and rectilinear, with an initial speed v 0 x.

Conditions for movement along the O X axis:

x 0 = 0 ; v 0 x = v 0 cos α ; a x = 0 .

Conditions for movement along the O Y axis:

y 0 = 0 ; v 0 y = v 0 sin α ; a y = - g .

Let us give formulas for the motion of a body thrown at an angle to the horizontal.

Body flight time:

t = 2 v 0 sin α g .

Body flight range:

L = v 0 2 sin 2 α g .

Maximum flight range is achieved at angle α = 45°.

L m a x = v 0 2 g .

Maximum lift height:

h = v 0 2 sin 2 α 2 g .

Note that in real conditions, the movement of a body thrown at an angle to the horizon can take place along a trajectory different from parabolic due to air and wind resistance. The study of the movement of bodies thrown in space is a special science - ballistics.

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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 towards the opposite side(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.

Movement of a body thrown at an angle to the horizontal

Basic formulas for curvilinear motion

1 . Speed of movement of a material point

\(\vec V=\frac(d\vec r)(dt)\) ,

where \(\vec r\) is the radius vector of the point.

2 . Acceleration of a material point

\(\vec a=\frac(d\vec V)(dt)=\frac(d^2\vec r)(dt^2)\),

\(a=\sqrt(a^2_(\tau)+a^2_n)\) ,

where \(a_(\tau)\) is tangential acceleration, \(a_n\) is normal acceleration.

3 . Tangential acceleration

\(a_(\tau)=\frac(dV)(dt)=\frac(d^2s)(dt^2)\)

4 . Normal acceleration

\(a_n=\frac(V^2)(R)\) ,

where \(R\) is the radius of curvature of the trajectory.

5 . for uniform motion

\(S=V_0t+\frac(at^2)(2)\)

\(V=V_0+at\)

Expressing \(t\) from the second equality and substituting it into the first, we obtain the useful formula

\(2aS=V^2-V_0^2\)

Examples of problem solving

In problems about the movement of a body in a gravity field, we will assume \(a=g=9.8\) m/s 2 .

Task 1.

The projectile flies out of the gun with an initial speed of 490 m/s at an angle of 30 0 to the horizontal. Find the height, range and time of flight of the projectile, without taking into account its rotation and air resistance.

Problem solution

Find: \(h, S, t\)

\(V_0=490\) m/s

\(\alpha=30^0\)

Let's connect the ISO with the gun.

The velocity components along the Ox and Oy axes at the initial time are equal:

\(V_(0x)=V_0\cos\alpha\) - remains unchanged throughout the flight of the projectile,

\(V_(0y)=V_0\sin\alpha\) - changes according to the equation of uniform motion

\(V_y=V_0\sin\alpha-gt\) .

At the highest point of rise \(V_y=V_0\sin\alpha-gt_1=0\) , from where

\(t_1=\frac(V_0\sin\alpha)(g)\)

Total projectile flight time

\(t=2t_1=\frac(2V_0\sin\alpha)(g)=50\) c.

We determine the height of the projectile from the formula for the path of equal slow motion

\(h=V_(0y)t_1-\frac(gt_1^2)(2)=\frac(V_0^2\sin^2\alpha)(2g)=3060\) m.

Let's define the flight range as

\(S=V_(0x)t=\frac(V_0^2\sin(2\alpha))(g)=21000\) m.

Problem 2.

A body falls freely from point A. At the same time, another body is thrown from point B at an angle \(\alpha\) to the horizon so that both bodies collide in the air. Show that the angle \(\alpha\) does not depend on the initial speed \(V_0\) of the body thrown from point B, and determine this angle if \(\frac(H)(S)=\sqrt3\) . Neglect air resistance.

Solving the problem.

Find: \(\alpha\)

Given: \(\frac(H)(S)=\sqrt3\)

Let's connect the ISO with point B.

Both bodies can meet on line OA (see figure) at point C. Let us decompose the speed \(V_0\) of a body thrown from point B into horizontal and vertical components:

\(V_(0x)=V_0\cos\alpha\) ; \(V_(0y)=V_0\sin\alpha\) .

Let time pass from the start of the movement to the moment of meeting

\(t=\frac(S)(V_(0x))=\frac(S)(V_0\cos\alpha)\).

During this time, the body from point A will fall by an amount

\(H-h=\frac(gt^2)(2)\) ,

and the body from point B will rise to a height

\(h=V_(0y)t-\frac(gt^2)(2)=V_0\sin\alpha(t)-\frac(gt^2)(2)\).

Solving the last two equations together, we find

\(H=V_0\sin\alpha(t)\) .

Substituting the previously found time here, we get

\(\tan\alpha=\frac(H)(S)=\sqrt3\),

those. The throwing angle does not depend on the initial speed.

\(\alpha=60^0\)

Task 3.

A body is thrown from a tower in a horizontal direction at a speed of 40 m/s. What is the speed of the body 3 s after the start of movement? What angle does the body’s velocity vector form with the horizontal plane at this moment?

Solving the problem.

Find: \(\alpha\)

Given: \(V_0=40\) m/s. \(t=3\) c.

Let's connect the ISO with the tower.

The body simultaneously participates in two movements: uniformly in the horizontal direction with a speed \(V_0\) and in free fall with a speed \(V_y=gt\) . Then the total speed of the body is

\(V=\sqrt(V_0^2+g^2t^2)=50 m/s.\)

The direction of the velocity vector is determined by the angle \(\alpha\) . From the figure we see that

\(\cos\alpha=\frac(V_0)(V)=\frac(V_0)(\sqrt(V_0^2+g^2t^2))=0.8\)

\(\alpha=37^0\)

Task 4.

Two bodies are thrown vertically upward from one point, one after the other, with a time interval equal to \(\Delta(t)\), with the same speeds \(V_0\) . After what time \(t\) after throwing the first body will they meet?

Solving the problem.

Find: \(t\)

Given: \(V_0\) , \(\Delta(t)\)

From the analysis of the problem conditions, it is clear that the first body will rise to the maximum height and on the descent will meet the second body. Let us write down the laws of motion of bodies:

\(h_1=V_0t-\frac(gt^2)(2)\)

\(h_2=V_0(t-\Delta(t))-\frac(g(t-\Delta(t))^2)(2)\).

At the moment of meeting \(h_1=h_2\), from where we immediately get

\(t=\frac(V_0)(g)+\frac(\Delta(t))(2)\)

Below are the conditions of the problems and scanned solutions. If you need to solve a problem on this topic, you can find a similar condition here and solve yours by analogy. The page may take some time to load due to a large number drawings. If you need problem solving or online help in physics, please contact us, we will be happy to help.

The principle of solving these problems is to decompose the speed of a freely falling body into two components - horizontal and vertical. The horizontal component of the velocity is constant, the vertical movement occurs with the acceleration of free fall g=9.8 m/s 2 . The law of conservation of mechanical energy can also be applied, according to which the sum of the potential and kinetic energy of the body in this case is constant.

A material point is thrown at an angle to the horizon with an initial speed of 15 m/s. The initial kinetic energy is 3 times greater than the kinetic energy of the point at the top point of the trajectory. How high did the point rise?

A body is thrown at an angle of 40 degrees to the horizontal with an initial speed of 10 m/s. Find the distance that the body will fly before falling, the height of rise at the top point of the trajectory and the time in flight.

A body is thrown down from a tower of height H, at an angle α to the horizontal, with an initial speed v. Find the distance from the tower to the place where the body fell.

A body with a mass of 0.5 kg is thrown from the surface of the Earth at an angle of 30 degrees to the horizontal, with an initial speed of 10 m/s. Find potential and kinetic energy body after 0.4 s.

A material point is thrown upward from the Earth's surface at an angle to the horizon with an initial speed of 10 m/s. Determine the speed of a point at a height of 3 m.

A body is thrown upward from the Earth's surface at an angle of 60 degrees with an initial speed of 10 m/s. Find the distance to the point of impact, the speed of the body at the point of impact and the time in flight.

A body is thrown upward at an angle to the horizontal with an initial speed of 20 m/s. The distance to the fall point is 4 times the maximum lift height. Find the angle at which the body is thrown.

A body is thrown from a height of 5 m at an angle of 30 degrees to the horizontal with an initial speed of 22 m/s. Find the flight range of the body and the time of flight of the body.

A body is thrown from the Earth's surface at an angle to the horizon with an initial speed of 30 m/s. Find tangential and normal acceleration bodies 1s after the throw.

A body is thrown from the surface of Zesli at an angle of 30 degrees to the horizontal with an initial speed of 14.7 m/s. Find the tangential and normal accelerations of the body 1.25 s after the throw.

A body is thrown at an angle of 60 degrees to the horizontal with an initial speed of 20 m/s. After what time will the angle between the speed and the horizon become 45 degrees?

Ball thrown in the gym at an angle to the horizon,with an initial speed of 20 m/s, at the top point of the trajectory it touched the ceiling at a height of 8 m and fell at some distance from the place of the throw. Find this distance and the angle at which the body is thrown.

A body thrown from the surface of the Earth at an angle to the horizon fell after 2.2 s. Find the maximum lifting height of the body.

A stone is thrown at an angle of 30 degrees to the horizontal. The stone reached a certain height twice - 1 s and 3 s after being thrown. Find this height and the initial speed of the stone.

A stone is thrown at an angle of 30 degrees to the horizontal with an initial speed of 10 m/s. Find the distance from the throwing point to the stone after 4 s.

The projectile is fired at the moment when the plane flies over the gun, at an angle to the horizon with an initial speed of 500 m/s. The shell hit the plane at an altitude of 3.5 km 10 seconds after being fired. What is the speed of the plane?

A cannonball with a mass of 5 kg is thrown from the surface of the Earth at an angle of 60 degrees to the horizontal. The energy spent to accelerate the weight is 500 J. Determine the flight range and flight time.

A body is thrown down from a height of 100 m at an angle of 30 degrees to the horizontal with an initial speed of 5 m/s. Find the flight range of the body.

A body with a mass of 200 g, thrown from the surface of the Earth at an angle to the horizon, fell at a distance of 5 m after a time of 1.2 s. Find a body throwing job.

Movement of a body thrown at an angle to the horizontal

Let us consider the movement of a body thrown with a speed V 0, the vector of which is directed at an angle α to the horizon, in XOY plane, positioning the body at the moment of throwing at the origin, as shown in Figure 1.

In the absence of resistance forces, the movement of a body thrown at an angle to the horizontal can be considered as special case curvilinear movement under the influence of gravity. Applying Newton's 2nd law


∑ F i

we get
mg = ma,


	a = g
The projections of the acceleration vector a on the OX and OU axes are equal:


		= −g
where g = const is	acceleration of gravity,			which is always
directed vertically downwards	numerical value g = 9.8 m/s2;		= −g		because op-amp axis on

Figure 1 is directed upwards, in the case when the OY axis is directed downward, then the projection of the vector

2 a on the op-amp axis will be positive(reading the conditions of the problems, choose the direction of the axes yourself, if this is not stated in the conditions).

The values of the projections of the acceleration vector a on the OX and OU axes give reason to make

the following output:

∙ a body thrown at an angle to the horizontal simultaneously participates in two movements - uniform horizontally and uniformly variable along


verticals.
The speed of the body in this case

	V = Vx + Vy

The speed of the body at the initial moment of time (at the moment of throwing the body)
V 0 = V 0 x			V 0 y .
The projections of the initial velocity vector on the OX and OU axes are equal
		Vcosα

V 0 y		V 0 sin α

For uniformly variable motion, the dependences of speed and displacement on time are given by the equations:

V 0 + at

S 0 + V 0 t +

and S 0 is the speed and displacement of the body at the initial moment of time,

and S t is the speed and displacement of the body at time t.

The projections of the vector equation (8) on the OX and OU axes are equal

V 0 x

Axt,

V ty = V 0 y + a y t

Const

V 0 y - gt

The projections of the vector equation (9) on the OX and OU axes are equal

S ox + V ox t +

a y t 2

S 0 y

Voy t +

taking into account equalities (4), we obtain

S 0 y

Voy t -

gt 2

where Sox and Soy are

body coordinates

at the initial moment of time,

and Stx and Sty -

coordinates of the body at time t.

During its movement t (from the moment of throwing to the moment of falling on the same

level) the body rises to the maximum height hmax, descends from it and flies away from the throwing point at a distance L (flight range) - see Figure 1.

1) Body movement time t can be found taking into account the values of body coordinates Sy in

Soy = 0, Sty = 0,

Substituting the values of Voy and (14) into the second equation of system (13), we obtain

2) Flight range L can be found, taking into account the values of the body coordinates Sх in

initial time and at time t (see Fig. 1)

Soх = 0, Stх = L,

Substituting the values of Vox and (17) into the first equation of system (13), we obtain

		L = V 0 cosα × t,
whence, taking into account (16), we obtain
L = Vcosα ×	2V sin α
L = Vcosα ×

3) Maximum lifting height h max can be found given the value

body velocity V at the point of maximum lift of the body

V 0 x

Because at this point V y

Using the second equations of systems (11) and (13),

the value of Voу, as well as the fact

that at the point of maximum lift of the body Sy = hmax, we obtain

0 = V 0 sin α - g × t under

gt sub2

V 0 sin α × t -

hmax

where tpod - rise time - time of movement to the height of maximum lift of the body.

Solving this system, we get

t under =

V 0 sin α

sin 2 α

Comparison of values (16) and (22) gives grounds to conclude

· time of movement to the height of maximum body lift (t under ) is equal to the time of descent of the body (tп) from this height and is equal to half the time of the entire movement of the body from the moment of throwing to the moment of falling to the same level

t under	Tsp

Studying the motion of a body thrown with a speed V 0, the vector of which is directed at an angle α to the horizontal, in the XOY plane, is very clear on a computer model

"Free fall of bodies" in the collection of computer models "Open Physics"

PHYSIKON company. In this model, you can set different initial conditions.

For example, the case we considered must be specified (the “Clear” command) with the initial condition h = 0 and selected V0 and α. The "Start" command will demonstrate the movement of the body and give a picture of the trajectory of movement and the direction of the body's velocity vectors at fixed moments in time.

Fig.2. Dialog window of the computer model "Free fall of bodies" in the section

"Mechanics"; a body moves from the origin and falls at the same level.

If the condition of the problem differs from the case we considered, then it is necessary

to solve the problem, choosing the direction of the axes, place the body at the initial moment

time, depict the trajectory of the body to the point of fall, thus

by determining the coordinates of the body at the initial and final moments of time. Then

use equations (3), (5), (8) and (9) as a basis for the solution and discussed above

algorithm for solving the problem.

Let's consider special cases.

6 1. The body was thrown at speed V 0 , whose vector is directed at an angleα to

horizon, from a height h and it fell at a distance L from the point of throwing. y to initial

Soy = h,

and the values of the remaining coordinates will be selected in the same way as we selected.

Fig.3. Dialog window of the computer model "Free fall of bodies" in the section

"Mechanics"; the body moves from point h = 50m and falls to zero level.

2. A body was thrown horizontally with a speed V 0 from a height h and it fell at a distance L from the point of throwing. The difference from the case we considered is that the values of the body coordinates S y at the initial moment will also be determined by equation (25),

and the values of the remaining coordinates will be selected in the same way as we selected. But in this case, the initial velocity of the body in projection onto the OU axis is equal to zero (since α = 0), i.e.

the projections of the initial velocity vector on the OX and OU axes are equal