Throw a ball up at one initial velocity and return to the throwing point at another velocity to find

1.First find the height that the ball can rise without resistance, 1 2mv square = mgh to get h, yes.

2.Then calculate the energy lost in the initial velocity and the final velocity: 1 2 mv first square minus 1 2 mv final square.

3.The drag force is f, so F multiplied by 2 h to get the resistance work, which is the kinetic energy lost by the ball, and the kinetic energy lost has been found in the second step, right.

In this way, the average resistance of the whole process is obtained, and if you want a certain point of resistance, it is more complicated to say, and I guess you will not be doing such a troublesome problem.

In addition: upstairs, there is resistance, time t is not easy to find, using the principle of conservation of energy is the simplest, most direct and the least wrong.

I think it's easier to use the kinetic energy theorem. Set the initial velocity to v0 and the final velocity to v1

Suppose the ball rises at a height of hThen in the process of ascent, by the kinetic energy theorem.

fh-mgh=0-mv0"2 (Note:.)"Represents square) in the whole process, the work done by gravity is 0, and the drag force always does negative work during the ascent and descent. By the kinetic energy theorem.

0-2fh=mv1"/2-mv0"/2

Simultaneous two equations, the only unknowns are f and hSo find f

Let the resistance be f and the height of the elevation be h

The square of v0 is not easy to express, it is written as v0*v0).

then by the law of conservation of momentum.

During the ascent:

f*h+mg*h=(1/2)mv0*v0

During the descent:

f*h+(1/2)mv1*v1=mg*h

Two equations, with only two unknowns, f and h

It can be found.

Let the velocity at the beginning be v1, the velocity of falling back to the parabola is v2, the mass of the ball is m, the drag is f, and the motion time of the ball is t.

Then the change in momentum of the ball is m(v1-v2).

The change in momentum comes from the impulse with the resistance, so m(v1-v2)=ft so that as long as you know v1, v2, m, t, you can find f.

It is still easier to use the kinetic energy theorem.

【Answer】 15m

Solution] stipulates that the vertical upward direction is the positive direction, from throwing to meeting, the first ball motion time is t, then the second ball motion time is (t-2).

When the two balls meet, the displacement is the same:

The displacement of the first ball is x1=v0t-gt; 2, and the displacement of the second ball is x2=v0(t-2)-g(t-2) 2, so v0t-gt; 2=v0(t-2)-g(t-2) 2 is substituted into the data to obtain t=3s

At this time, the displacement (height from the throw point) of both is 15m

You can't just say that.

Add an assumption: the resistance is a constant force.

Then FH = loss of the same kinetic energy when falling.

v = root number (2e m) = root number 15 5 * vo

Let the initial velocity be v and the ascent height be h, which is known by the kinetic energy group grip theorem.

2kmgh=1 2m( (1) The whole process is withered or submerged.

kmgh-mgh=0-1 2mv 2 (2) ascending process.

1) (2) Solutionable k

The instantaneous velocity of the ball falling on both inclined planes, regardless of the initial velocity when the ball is thrown.

The ball is doing a flat throwing motion, and the speed of the flat throwing motion in the horizontal direction is constant, so the change in the velocity of the ball occurs in the vertical direction, and the change in the velocity in the vertical direction is v=g t, therefore, the velocity of the ball that has been in motion for a long time changes greatly, so the velocity of ball A changes the most, so B is correct;

The speed of the velocity change is the acceleration of the object's motion, because the object is doing a flat throwing motion, and the acceleration of the motion is the gravitational acceleration g, so the speed of the three motion changes is the same, so c is wrong;

First of all, it is impossible to be vertical at point A anyway, and then look at points b and c, the vertical speed is GT, the horizontal speed is v, and then the angle of the inclined plane is, to combine the vertical slope of the velocity, after combining the two velocities, you need =tan, that is, v=, Lu Zhengkai then when the time of t passes, the vertical displacement is, and the horizontal displacement is vt=(?).t= that is, in order to satisfy this relation, the horizontal displacement and vertical displacement need to be the same, obviously in the figure b and c is impossible to complete, because the horizontal displacement on b and c must be greater than the vertical displacement, so the instantaneous velocity falling on the two inclined planes cannot be perpendicular to the inclined plane, so d is correct

Yes, according to the law of gravitation, as long as the wax is under vacuum conditions, the speed allowance they fall on is the same, and if it is not for the vacuum, it may be different if it is not affected by the resistance of the air knowledge annihilation.

This problem is solved using conservation of energy. Assuming that the air resistance is constant during the rise and fall, the work of overcoming the air resistance is also set to w. Old.

1, 2, one-way is mgh =

The result is: h=

As the object is subjected to gravity and drag;

a=（mg+f)/m

From v 2 = 2ax we get:

h=mv0^2/2(mg+f)

Since resistance is always doing negative work; So from the throwing point to the back to the throwing point, the kinetic energy theorem gets:

2FH=1 2MV 2-1 2MV0 2 solution: V=V0 2-2FV0 2 (mg+h) under the root number

It comes out directly according to the conservation of energy! The ball rises in height as follows: m(v0) squared divided by 2 times (f+mg) and falls back at a velocity of (v0 squared minus (4hf divided by m)).

The ball initially has a kinetic energy of 1 2mv0 2

After reaching the highest point, it is converted into gravitational potential energy mgh and heat energy FH, that is, 1 2mv0 2=mgh+fh

Therefore, after h=(1 2mv0 2) (mg+f) falls from the highest point, the gravitational potential energy mgh is converted into 1 2mv 2 and heat energy fh, that is, mgh=1 2mv 2+fh

From this we can get v.