Eager to throw vertically in high school physics, and throw vertically in high school physics questi

The ascent phase is a uniform deceleration movement, and the final velocity is 0: 20-gt=0 g=10, t=2 seconds.

Then the falling distance in the last 2s is 100m: 100=v*2+1 2*g*2 2 The velocity of the penultimate second is obtained: 40m s

Then from this speed, you can column: 40 = GT to find t = 4 seconds, so far when the rope of this object is broken, it rises to the highest point after 2 seconds, and then descends to 100 meters above the ground after 4 seconds, and lands after 2 seconds.

The whole process is 8 seconds. As for the distance between the object and the ground when the rope breaks: subtract the distance of falling by 2 seconds by the distance of the object falling for 6 seconds, and the former is the highest height at which the object rises.

h6 = 1 2g * 6 2 = 180m, h2 = 1 2g * 2 2 = 20m, and the final solution is 160m.

Let the velocity of the last 2s money be v vt+1 2gt 2=100(t=2) v=40m s 2gh=v 2 h=80m h1=v0 2 2g=20m h=h+100-h160m second question v0=gt1 , t1=2s ; v=gt2 , t2=4s ;t=t1+t2+t=8s where t2+t is the time taken to descend and t1 is the time to rise.

Understanding the movement of the heavy object is the key: the heavy object does an upward uniform deceleration motion with a muzzle velocity of 20m s.

The average speed of the last 2 seconds v=100 2=50 (m s) The time required to fall from the highest point t t = v g + 1 = 6 (s) The height of the weight from the ground when the rope is broken: h =

t=6+2=8(s)

The movement process is to decelerate and move upwards until the speed is 0 and then move freely, and the free fall motion moves 100 meters in the last 2 seconds, and the initial velocity of the known time acceleration and distance is obtained by 40m s (that is, the speed of the last 2s), that is, when the object lands, the speed is 40 + 10 * 2 = 60m s, the deceleration moves upwards by 20m, and the free fall moves by 180m, so the ground clearance is 160m and the time is 8 seconds.

The height is 180m, let the velocity of the last 2s money be v vt+1 2gt 2=100(t=2) v=40m s 2gh=v 2 h=80m h=h+100=180m The second question v0=gt1 , t1=2s ; v=gt2 , t2=4s ;t=t1+t2+t=8s where t2+t is the time taken to descend and t1 is the time to rise.

v0-gt= ±v0/3t1=2v0/(3g)

t2=4v0/(3g)

Benson holds the pre-question of this clear skin wild selection b, d

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Problem description:1Throw an object vertically on the ground and pass through the window of the upstairs with a height of meters in seconds. When the object falls back. The time it takes to fall from the window to the ground is seconds. Find the maximum height reached by the object.

2.A balloon rises straight up from the ground at a uniform velocity of 5 meters per second, and after 30 seconds, a heavy object falls. Ask (1) How long does it take for an object to fall to the ground? (2) The velocity of the object reaching the ground. （g=10m/s2）

Analysis: Question 1:

Solution: s=vt-gt; 2 (where v represents the velocity of the object before passing through the window).

s=substitution:

v=So, the height of the bottom of the window to the highest point is s1=v 2g=

Since the object is only subjected to gravity during its rise and fall, when the object falls back from the highest point to the bottom of the window, its velocity is also v, i.e., by the title.

s=vt1+gt1 2 (s is the height from the bottom of the window to the ground).

t1=；v=

s=s=So, the maximum height reached by an object h=Question 2:

Solution: At the end of 30 seconds, the balloon reaches the same height as the weight.

30×5=150m

The velocity of the weight when it leaves the balloon v=5m s

When the velocity of the heavy object becomes 0 (i.e., the velocity of the heavy object rising to the maximum height is 0), that is, the height at which the heavy object rises again after leaving the balloon is h1=v 2g=25 20=

The time taken for the ascent is v g=5 10=

The maximum height of a heavy object ascent is 150+

s = gt 2 (t is the fall time).

t = t = So, the object has to pass through ( before it can fall to the ground, i.e. 6s

Let the velocity of the object reach the ground be v2 according to the time t= of the object falling from the highest point to the ground in (1).

Get: v2=gt=10

So, the velocity of the object reaching the ground is 55m s

Vertical upward throwing motion.

It is an important part of high school physics, and the following is a summary of the formula of vertical upward throwing motion in high school physics that I bring to you, I hope it will be helpful to you.

1.Displacement s=vot-gt2 2

2.Final velocity vt=vo-gt(g=

3.Inference vt2-vo2=-2gs

4.The maximum height of the rise hm=VO2 2G (throw the point from Lusun Yuan).

5.Round-trip time t = 2vo g (time from throwing back to position).

Note: 1) The whole process processing: it is a uniform deceleration linear motion, with upward as the positive direction, and acceleration.

Take a negative value; 2) Segmented treatment: upward is a uniform deceleration linear motion, downward is a free fall motion.

has symmetry;

3) The process of ascent and fall is symmetrical, such as the equivalent reversal of velocity at the same point.

Understand. High school students should take the initiative to listen to the lectures, listen to every word the teacher says, and memorize the definition of high school physics concepts.

Keep in mind. Especially the basic concepts. Definitions, laws, conclusions, etc., don't regard these companion states as knowledge that can be remembered or not, and if you ignore them, high school students' understanding and application of physical problems will be hindered, and they will lose points due to unclear concepts in the process of physical problem solving

The basic concepts are clear, the basic rules are familiar, and the basic methods are all categories to be remembered. Only in this way will high school students be able to learn physics and solve various problems.

will use. The ability to use is the foundation of improving performance, that is, to master the concepts, formulas, etc. flexibly, learn and use them lively, not rote memorization, different question types use different solution methods, and the use of formulas is also flexible and changeable, so as to achieve the purpose of correct problem solving. For example, for Newton's three laws of motion.

What is momentum and why momentum is conserved The understanding of these basic concepts of dynamics is boring and even difficult to understand just by staying literally, and these knowledge affects the entire learning process of mechanics.

Practiced.

1(1)v end 2-v beginning 2 2a=s

The solution is s=5 meters (note that the upward throwing motion is done, g is a negative value).

2) v at the beginning - at = v at the end.

i.e. 10-10t=0

The solution yields t=13)v-2-v-2 2a=s

i.e. v end 2-0 (2*10)=5

The solution yields v end = -10m s

2(1)The beginning of v + AT=the end of V.

That is, 40-10t=0

The solution of t=4 is equal from the time it is thrown to the time it returns to its original position. That is, the time from the time of the object to the time from the throw to the point back to the throw point is 2t = 8

Again. v at the beginning t t + 1 2at 2 = s (v at the beginning refers to the velocity at which the object will throw the point when it moves) i.e.

40*t+5t^2=10

The solution yields t = 3, root numbers 2-4

t = -4-3 root number 2 (rounded), so the time of the object landing t=8 + 3 root number 2-4 = 4 plus 3 root number 2

2) The speed at which the object lands can be seen as the speed at which the object lands after moving to the highest point.

The time from the highest point to the landing is t=4+3, the root number is 2-4=3, the root number is 2

So got. v at the beginning + at = v at the end.

Namely. 0 + 10 * 3 root number 2 = v end.

The solution gives v end = 30 root number 2

3 (1) The chain state shows that the object can move upwards for 1 s. at the beginning of 30m

1/2gt^2=s

i.e. 1 2 * 10 * t 2 = 30

The solution is t = root number 6

So the time of the upward movement of the object t = t + 1 = root number 6 + 1

s=1 2gt; 2=1 2*10*(root number 6+1) 2=35+10 root number 6

Shed Qingyuan 2) v end = v beginning + gt = 0 + 10 * (root number 6 + 1) = 10 + 10 root number 6

When the vertical upward throw returns to the origin, the velocity is the same as that of the vertical downward toss.

is the initial velocity. The height at which it is thrown.

h=vt+g*t^2*1/2=35+10*1/2=40m

The main difference between these two processes is that there is more upward and downward process than downward toss, and the time difference of this process can be (8-1)=7s according to the title

Therefore, if the initial velocity is v=gt, the rise and fall to the same point are actually inverse processes to each other, so the time is the same, so v=10*

h=vt+1 2gt 2, and the solution is h=40m

So the answer is that the height is 40 meters and the speed is 35

Find the maximum height first.

v0 square = 2gh

h=v0 square 2g

It can be known that 1 2 height, v0 square 4g

Then find the time. Use the formula s=v0t+1 2gt square.

Pay attention to the positive and negative signs of the direction and calculate the results yourself.

In fact, there is still a big trap in this question!

v end = v beginning - gt = 30 - 10t = 0

We get t=3s, that is, in three seconds, the object has reached its highest point and the velocity is 0So at the fifth second, the velocity is -20 meters per second (the specified upward direction is positive).

From x=v initial t-1 2gt 2 can bring in the data, the resulting displacement is 25 meters. The direction of displacement is upward. The amount of change in velocity is -50 meters per second, and the average velocity is 5 meters per second, in an upward direction.