Two small balls of A and B with different masses are at the same height, and ball A is thrown vertic

First of all, the motion of free fall is not related to mass in the absence of resistance.

Secondly, looking at A, from the symmetry of the upward throwing motion, when A falls back to the original height, the velocity direction is downward, and the magnitude is v0, then its next motion is no different from B, so they reach the water surface at the same speed.

The gravitational potential energy of the transformation is proportional to m, and the kinetic energy is also proportional to m, so the velocity is the same.

Because A and B fall from the same height to the same height, ignoring air resistance and all other resistance, only gravity does work, gravity is a conservative force, and the work done is related to the initial position, so...

Specifically, it is as follows:

mg*h = work done by gravity = increment of kinetic energy, v2 is the square of v. (v0)2 is the square of v0.

Solve v is a constant, and there is no unknown quantity in it, so choose d

The more vivid statement that can be used as a comparison is that when A throws up and falls down to his initial height, the velocity is exactly the same as B's initial velocity, and A begins to carry out the same process as B's movement, so the velocity is the same when it falls to the surface of the water.

When he throws the ball, only gravity does the work, and since the height is equal, and the kinetic energy theorem is known, the velocity of them when they fall into the water is the same.

The law of conservation of mechanical energy.

If you're right, hurry up and give points! Don't drag.

The answer is Da, when landing, the vertical upward velocity should be larger, the vertical throwing speed = V+2H T0, where T0 is the time for the object to fall freely from the initial height to the ground, and the horizontal throwing landing speed is [V 2+(2H T0)].

b, when landing, the instantaneous velocity of gravity is not the same, the gravity velocity of vertical upward throwing is v+2h t0, and the gravity landing speed of horizontal throwing is 2h t0, so the instantaneous power is not the same.

c, gravity does work, the change of gravitational potential energy and the change of kinetic energy in the vertical direction, so the gravitational force thrown vertically upwards does more work.

d, from the beginning to the landing, the acceleration of the two balls is the same, but the direction of the speed of the two balls is different, the resultant force is not the same, the size of the side banquet is different, the vertical size change is GT, and the speed of the flat throw changes all the time.

Let the velocity of ball A relative to ball B be v, the velocity of ball A after collision is v1, ball B is the mass of ball A m1, and ball B is m2

m1v=m1v1+m2v2 (Momentum is conserved.

1 2 m1v 2 = 1 2 m1v1 2 + 1 2m2v2 2 (Conservation of kinetic energy.)

Lianlead v1 = (m1-m2)v (m1+m2)v2=2m1v (m1+m2).

Because m1 = m2

So v1=0

v2=v because the direction of v is the direction of the original motion.

So the velocity of the ball is not 0

The velocity of the ball is 0

The answer is a

m1 won't be equal to m2 ah, so how can v1 be equal to 0

C test question analysis: because the two balls are exactly the same, and are only affected by gravity, so the mechanical energy of the two balls is conserved, then when landing, the speed of the two balls is the same, but the direction is different, a is wrong, according to the conservation of mechanical energy, the speed of the two balls is the same when landing in the balance section, but the direction of their speed is different, the direction of the velocity of the ball thrown vertically is vertically downward, but the ball thrown flat falls.

The velocity of ground time has a certain angle with the horizontal direction, so when the two small balls land, the instantaneous power of gravity.

is the size of the angle between the velocity and the vertical direction) is different, so b is wrong; The magnitude of the work done by gravity is only related to the position of the object at the beginning and end, two balls are thrown from the same place, and finally they both fall to the ground in defeat, and the difference in height between them is the same, so gravity does the same work on the two balls, so c is correct; Gravity does the same amount of work for the two balls, but they move for different times, so the average power of gravity is the same.

different, so d wrong;

Therefore, choose C Comment: When answering this question, you must pay attention to the difference in the calculation methods of average power and instantaneous power

A. In the process of motion, the two small balls only have gravity to do work, and the mechanical energy is conserved, so according to the conservation of mechanical energy, it can be known that the velocity of the two objects is equal when they land, and the direction is different, so the speed is different, so A is wrong

B. When reaching the bottom end, the velocity of the two objects is the same, and the gravity is also the same, but the gravity of object A has an angle with the velocity, and the gravity of object B is the same as the direction of velocity, so the instantaneous power of the gravity of object B before landing is greater than the instantaneous power of the gravity of object A, so B is wrong

C. According to the expression of the work done by gravity, the work of the two balls in the process of motion is known to be mgh by gravity, so C is correct

D. From the beginning of the movement to the landing, gravity does the same work on the two small balls, but the time required for process A is less than the time required for B, according to P=Wt

Knowing that the average power of gravity to do work on the two small balls is not the same, d is wrong;

So choose A

If the choice is correct, then it is a

BC because when throwing, v A v B, and the mass of A and B is equal, so A is right.

However, because there is no other force to do work in the whole process except gravity, the mechanical energy is conserved, and because the gravitational potential energy of the two is the same at the beginning, but the kinetic energy A is greater, so B is right, and then because at the highest point, its mechanical energy is only gravitational potential energy, and the mechanical energy of A is greater at the beginning and end, so C is right, and finally, when returning to the throwing point, according to the conservation of mechanical energy, the kinetic energy of A and B is equal to the kinetic energy of the cover when it is thrown, so D is wrong.

The acceleration of a, b, d, and vertical upward throwing motion is the same, according to the velocity symmetry of the vertical upward throwing motion, it can be known that when ball A returns to the throwing point, the velocity is v0, and then the vertical downward throwing motion is done, so the velocity is the same as the speed of B when it reaches the water surface, so a is wrong, b is also wrong, and d is correct;

c. The acceleration of the vertical upward throwing motion is the same, so the speed of the two balls reaching the water surface has nothing to do with the mass, so C is wrong;

Therefore, choose D

Kinetic energy = 1 2mV2

Because both balls are equal.

So the first cell =

Second, the third cell kinetic energy is converted into gravitational potential energy.

The kinetic energy is the smallest, and the gravitational potential energy is the highest (i.e., the highest point).

So the fourth cell is 0

The fifth and sixth cells are converted into kinetic energy by gravitational potential energy.

=；Kinetic energy; gravitational potential energy; 0；gravitational potential energy; Kinetic energy;

Is B changed to A?

The velocity of A after ascending t=v g is 0, and the rise of s=v 2 2g, and the falling height of B in this time period is also s=v 2 2g, and the velocity of B at time t is downward in the direction of v.

The condition for the upward stage to meet is: h<2s=v 2 g

The falling section collides, A and B need to touch each other within the next T time, otherwise A does not touch after the initial position, and do not touch in the future, and the condition for H-2S collision is 0

b.From the time of throwing to the time of landing, the average power of gravity to do work on the two balls is equal.

d.When landing, the instantaneous power of gravity to do work on the two balls is equal.