If the body is thrown vertically upwards with an initial velocity υ 0 , it will move uniformly with an acceleration equal to a = - g = - 9 , 81 υ c 2 .
Picture 1
The tossing height h in time t and the speed υ through the interval t can be determined by the formulas:
t m a x is the time during which the body reaches its maximum height h m a x \u003d h, at υ \u003d 0, and the height h m a x itself can be determined using the formulas:
When the body reaches a height equal to h m a x , then it has a speed υ = 0 and an acceleration g . It follows that the body will not be able to remain at this height, so it will go into a state of free fall. That is, a body thrown upwards is a uniformly slow motion, in which, after reaching h m a x, the signs of displacement change to the opposite ones. It is important to know what was the initial height of the movement h 0 . The total time of the body will take the designation t, the time of free fall - t p, the final speed υ k, from here we get:
If the body is thrown vertically upward from ground level, then h 0 = 0.
The time required for a body to fall from a height where the body was previously thrown is equal to the time it takes to rise to its maximum height.
Since at the highest point the speed is zero, it can be seen:
The final velocity υ k of a body thrown vertically upwards from ground level is equal to the initial velocity υ 0 in magnitude and opposite in direction, as shown in the graph below.
Picture 2
Examples of problem solving
Example 1A body was thrown vertically upwards from a height of 25 meters at a speed of 15 m/s. How long will it take for it to reach the ground?
Given: υ 0 \u003d 15 m / s, h 0 \u003d 25 m, g \u003d 9, 8 m / s 2.
Find: t .
Decision
t \u003d υ 0 + υ 0 2 + g h 0 g \u003d 15 + 15 2 + 9, 8 25 9, 8 \u003d 3, 74 s
Answer: t = 3.74 s.
Example 2
A stone was thrown from a height h = 4 vertically upwards. Its initial speed is υ 0 = 10 m / s. Find the height to which the stone can rise to the maximum, its flight time and the speed with which it reaches the surface of the earth, the path traveled by the body.
By itself, the body does not move upward, as is known. It needs to be “thrown”, i.e., to inform it of some initial speed directed vertically upwards.
A body thrown upward moves, as experience shows, with the same acceleration as a freely falling body. This acceleration is equal and directed vertically downwards. The motion of a body thrown upwards is also a rectilinear uniformly accelerated motion, and the formulas that were written for the free fall of a body are also suitable for describing the motion of a body thrown upwards. But when writing formulas, one must take into account that the acceleration vector is directed against the initial velocity vector: the absolute value of the body's velocity does not increase, but decreases. Therefore, if the coordinate axis is directed upwards, the projection of the initial velocity will be positive, and the projection of acceleration will be negative, and the formulas will take the form:
Since a body thrown up moves with decreasing speed, there will come a moment when the speed becomes zero. At this point, the body will be at its maximum height. Substituting the value into formula (1) we get:
From here you can find the time it takes for the body to rise to its maximum height:
The maximum height is determined from formula (2).
Substituting into the formula we get
After the body reaches a height, it will begin to fall down; the projection of its speed will become negative, and will increase in absolute value (see formula 1), while the height will decrease with time according to formula (2) at
Using formulas (1) and (2), it is easy to make sure that the speed of the body at the moment of its fall to the ground or in general to where it was thrown from (at h = 0) is equal in absolute value to the initial speed and the time of the fall of the body is equal to the time of his rise.
The fall of a body can also be considered separately as a free fall of a body from a height. Then we can use the formulas given in the previous paragraph.
Task. A body is thrown vertically upward with a speed of 25 m/sec. What is the speed of the body after 4 seconds? What movement will the body make and what is the length of the path traveled by the body during this time? Decision. The speed of the body is calculated by the formula
By the end of the fourth second
The sign means that the speed is directed against the coordinate axis directed upwards, i.e., at the end of the fourth second, the body was already moving downwards, having passed through the highest point of its ascent.
The amount of displacement of the body is found by the formula
This movement is counted from the place where the body was thrown from. But at that moment the body was already moving down. Therefore, the length of the path traveled by the body is equal to the maximum height of the ascent plus the distance by which it managed to go down:
The value is calculated by the formula
Substituting the values we get: sec
Exercise 13
1. An arrow is fired from a bow vertically upwards at a speed of 30 m/sec. How high will she rise?
2. A body thrown vertically upwards from the ground fell after 8 seconds. Find to what height it rose and what was its initial speed?
3. From a spring gun located at a height of 2 m above the ground, a ball flies vertically upwards at a speed of 5 m/sec. Determine to what maximum height it will rise and what speed the ball will have at the moment it falls to the ground. How long was the balloon in flight? What is its movement during the first 0.2 seconds of flight?
4. A body is thrown vertically upward at a speed of 40 m/s. At what height will it be in 3 and 5 seconds and what will its speed be? To accept
5 Two bodies are thrown vertically upwards with different initial velocities. One of them reached four times the height of the other. How many times was its initial velocity greater than the initial velocity of the other body?
6. A body thrown upwards flies past the window at a speed of 12 m/sec. With what speed will it fly past the same window down?
A freely falling body can move in a straight line or along a curved path. It depends on the initial conditions. Let's consider this in more detail.
Free fall without initial velocity (υ 0 = 0) (Fig. 1).
With the chosen coordinate system, the movement of the body is described by the equations:
\(~\upsilon_y = gt, y = \frac(gt^2)(2) .\)
From the last formula, you can find the time the body falls from a height h\[~t = \sqrt(\frac(2h)(g))\]. Substituting the found time into the formula for velocity, we obtain the modulus of the body's velocity at the moment of fall\[~\upsilon = \sqrt(2gh)\].
The motion of a body thrown vertically upwards with initial velocity\(~\vec \upsilon_0\) (Fig. 2).
The motion of the body is described by the equations:
\(~\upsilon_y = \upsilon_0 - gt, y = \upsilon_0 t - \frac(gt^2)(2) .\)
From the velocity equation, it can be seen that the body moves uniformly slow up, reaches its maximum height, and then moves uniformly accelerated down. Considering that at y = h max speed υ y = 0 and at the moment the body reaches its initial position y= 0, you can find\[~t_1 = \frac(\upsilon_0)(g)\] - the time the body rises to the maximum height;
\(~h_(max) = \frac(\upsilon^2_0)(2g)\) - maximum body lift height;
\(~t_2 = 2t_1 = \frac(2 \upsilon_0)(g)\) - body flight time;
\(~\upsilon_(2y) = -\upsilon_0\) - velocity projection at the moment the body reaches its initial position.
Literature
Aksenovich L. A. Physics in high school: Theory. Tasks. Tests: Proc. allowance for institutions providing general. environments, education / L. A. Aksenovich, N. N. Rakina, K. S. Farino; Ed. K. S. Farino. - Mn.: Adukatsia i vykhavanne, 2004. - S. 14-15.
This video tutorial is for self-study topic "Movement of a body thrown vertically upwards". During this lesson, students will gain an understanding of the motion of a body in free fall. The teacher will talk about the movement of a body thrown vertically upwards.
In the previous lesson, we considered the issue of the motion of a body that was in free fall. Recall that free fall(Fig. 1) we call this movement, which occurs under the influence of gravity. The force of gravity is directed vertically downward along the radius towards the center of the Earth, acceleration of gravity while equal to .
Rice. 1. Free fall
How will the movement of a body thrown vertically upwards differ? It will differ in that the initial velocity will be directed vertically upwards, i.e., it can also be considered along the radius, but not towards the center of the Earth, but, on the contrary, from the center of the Earth upwards (Fig. 2). But the acceleration of free fall, as you know, is directed vertically downwards. So, we can say the following: the movement of the body vertically upwards in the first part of the path will be slow motion, and this slow motion will also occur with free fall acceleration and also under the action of gravity.
Rice. 2 Movement of a body thrown vertically upwards
Let's turn to the figure and see how the vectors are directed and how it fits with the frame of reference.
Rice. 3. Movement of a body thrown vertically upwards
AT this case reference system is connected to the earth. Axis Oy is directed vertically upwards, as is the initial velocity vector. The downward force of gravity acts on the body, which imparts to the body the acceleration of free fall, which will also be directed downward.
The following thing may be noted: the body will move slow, will rise to a certain height, and then will start rapidly fall down.
We have designated the maximum height, while .
The movement of a body thrown vertically upwards occurs near the surface of the Earth, when the free fall acceleration can be considered constant (Fig. 4).
Rice. 4. Near the surface of the Earth
Let us turn to the equations that make it possible to determine the speed, instantaneous speed and the distance traveled during the considered movement. The first equation is the velocity equation: . The second equation is the equation of motion for uniformly accelerated motion: .
Rice. 5. Axis Oy pointing up
Consider the first frame of reference - the frame of reference associated with the Earth, the axis Oy directed vertically upwards (Fig. 5). The initial velocity is also directed vertically upwards. In the previous lesson, we already said that the acceleration of free fall is directed downward along the radius towards the center of the Earth. So, if we now reduce the velocity equation to a given reference frame, then we get the following: .
It is a projection of the speed at a certain point in time. The equation of motion in this case is: 
.
Rice. 6. Axis Oy pointing down
Consider another reference system, when the axis Oy directed vertically downwards (Fig. 6). What will change from this?
. The projection of the initial velocity will be with a minus sign, since its vector is directed upwards, and the axis of the selected reference system is directed downwards. In this case, the acceleration of free fall will be with a plus sign, because it is directed downwards. Motion equation: 
.
Another very important concept to consider is the concept of weightlessness.
Definition.Weightlessness- a state in which the body moves only under the influence of gravity.
Definition. The weight- the force with which the body acts on the support or suspension due to attraction to the Earth.
Rice. 7 Illustration for weight determination
If a body near the Earth or at a short distance from the Earth's surface moves only under the action of gravity, then it will not act on the support or suspension. This state is called weightlessness. Very often, weightlessness is confused with the concept of the absence of gravity. In this case, it must be remembered that weight is the action on the support, and weightlessness- this is when there is no effect on the support. Gravity is a force that always acts near the surface of the Earth. This power is the result gravitational interaction with the earth.
Let us pay attention to one more important point related to the free fall of bodies and the movement vertically upwards. When the body moves up and moves with acceleration (Fig. 8), an action occurs, leading to the fact that the force with which the body acts on the support exceeds the force of gravity. If this happens, this state of the body is called overload, or the body itself is said to be overloaded.
Rice. 8. Overload
Conclusion
The state of weightlessness, the state of overload - these are extreme cases. Basically, when a body is moving on a horizontal surface, the weight of the body and the force of gravity most often remain equal to each other.
Bibliography
- Kikoin I.K., Kikoin A.K. Physics: Proc. for 9 cells. avg. school - M.: Enlightenment, 1992. - 191 p.
- Sivukhin D.V. General course physics. - M .: State publishing house of technical
- theoretical literature, 2005. - T. 1. Mechanics. - S. 372.
- Sokolovich Yu.A., Bogdanova G.S. Physics: Handbook with examples of problem solving. - 2nd edition, redistribution. - X .: Vesta: Publishing house "Ranok", 2005. - 464 p.
- Internet portal "eduspb.com" ()
- Internet portal "physbook.ru" ()
- Internet portal "phscs.ru" ()
Homework
Task 10001
The body is thrown vertically upwards with an initial speed v 0 =4 m/s. When it has reached the top point of the flight from the same starting point, the second body is thrown vertically upwards with the same initial speed v 0 . At what distance h from the starting point will the bodies meet? Air resistance is ignored.
Problem 14412
A body is thrown vertically upwards with an initial velocity v 0 = 9.8 m/s. Plot the dependence of height h and speed v on time t for the interval 0 ≤ t ≤ 2 s after 0.2 s.
Problem 14513
A stone of mass m = 1 kg is thrown vertically upwards with an initial velocity v 0 = 9.8 m/s. Construct a graph of the dependence on time t of the kinetic W k, potential W p and total W stone energies for the interval 0 ≤ t ≤ 2 s.
Problem 13823
A body is thrown vertically upwards with an initial velocity of 30 m/s and reaches highest point lift after 2.5 s. What was the average value of the air resistance force acting on the body during the ascent? Body weight 40 g.
Problem 18988
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 15 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies started to move simultaneously and after a time t = 0.2 s the distance between them became h = 5 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 18990
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 20 m/s, body B falls from a height H = 5 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.1 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 18992
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 7.5 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.8 s the distance between them became equal to h = 16 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 18994
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 25 m / s, body B falls from a height H = 23 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.32 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 18996
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 12.5 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.24 s the distance between them became equal to h = 2 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 18998
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 22 m/s, body B falls from a height H = 21 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.5 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Task 19000
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 5 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies started to move simultaneously and after a time t = 1.4 s the distance between them became h = 7 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19002
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 6.25 m / s, body B falls from a height H = 6 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.8 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19004
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 25 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies started to move simultaneously and after a time t = 0.2 s the distance between them became h = 11 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19006
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 8 m / s, body B falls from a height H = 19 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 1.25 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19008
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 10 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies started to move simultaneously and after a time t = 0.7 s the distance between them became h = 3 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19010
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 12 m/s, body B falls from a height H = 17 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 1.0 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19012
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upwards with an initial speed v 01 = 20 m/s, body B falls from a height H with an initial speed v 02 = 0. The bodies started to move simultaneously and after a time t = 0.35 s the distance between them became h = 5 m. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19014
Bodies A and B move towards each other along the same vertical. Body A is thrown vertically upward with an initial speed v 01 = 12.5 m / s, body B falls from a height H = 9 m with an initial speed v 02 = 0. The bodies began to move simultaneously and after a time t = 0.4 s the distance between them became equal to h. Find H, t 1 . Determine the time after which the bodies will meet.
Problem 19390
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.5 kg thrown vertically upwards with an initial speed v 0 = 4.9 m/s, at times t 1 = 0.2 s and t 2 = 0.8 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19392
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.5 kg thrown vertically upwards with an initial speed v 0 = 4.9 m/s, at times t 1 = 0.4 s and t 2 = 0.6 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19394
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.2 kg thrown vertically upwards with an initial speed v 0 = 19.6 m / s, at times t 1 = 0.8 s and t 2 = 3.2 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19396
Calculate the values of the kinetic, potential and total energy of a body with mass m = 0.2 kg thrown vertically upwards with an initial speed v 0 = 19.6 m/s, at times t 1 = 1.6 s and t 2 = 2.4 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19398
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.4 kg thrown vertically upwards with an initial speed v 0 = 12.25 m/s, at times t 1 = 0.5 s and t 2 = 2 s. Plot graphs of kinetic, potential and total energy versus time.
Problem 19400
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.4 kg thrown vertically upwards with an initial speed v 0 = 12.25 m/s, at times t 1 = 1 s and t 2 = 1.5 s. Plot graphs of kinetic, potential and total energy versus time.
Problem 19402
Calculate the values of the kinetic, potential and total energy of a body with mass m = 0.6 kg thrown vertically upwards with an initial speed v 0 = 2.45 m/s, at times t 1 = 0.1 s and t 2 = 0.4 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19404
Calculate the values of the kinetic, potential and total energy of a body with mass m = 0.6 kg thrown vertically upwards with an initial speed v 0 = 2.45 m/s, at times t 1 = 0.2 s and t 2 = 0.3 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19406
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.3 kg, thrown vertically upwards with an initial speed v 0 = 14.7 m / s, at times t 1 = 0.6 s and t 2 = 2.4 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19408
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.3 kg, thrown vertically upwards with an initial speed v 0 = 14.7 m / s, at times t 1 = 1.2 s and t 2 = 1.8 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19410
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.25 kg, thrown vertically upwards with an initial speed v 0 = 9.8 m / s, at times t 1 = 0.4 s and t 2 = 1.6 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19412
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.25 kg, thrown vertically upwards with an initial speed v 0 = 9.8 m / s, at times t 1 = 0.8 s and t 2 = 1.2 with. Plot graphs of kinetic, potential and total energy versus time.
Problem 19414
Calculate the values of the kinetic, potential and total energy of a body with mass m = 0.1 kg thrown vertically upwards with an initial speed v 0 = 24.5 m/s, at times t 1 = 1 s and t 2 = 4 s. Plot graphs of kinetic, potential and total energy versus time.
Problem 19416
Calculate the values of the kinetic, potential and total energy of a body with a mass m = 0.1 kg thrown vertically upwards with an initial speed v 0 = 24.5 m/s, at times t 1 = 2 s and t 2 = 3 s. Plot graphs of kinetic, potential and total energy versus time.