Help explain a flat tossing problem

It's a matter of velocity synthesis vs. decomposition! Looking at the vertical direction alone, A and B are always on the same horizontal line (under the condition of the question). Therefore, we will meet!

Song of the End of the Heavens] is quite right, using the decomposition of motion (speed) to analyze this problem, it is very simple.

There is a premise in this question that is not stated, that is, the horizontal initial velocity of ball A is towards ball B - otherwise they can only get farther and farther away and never collide.

On this premise, there are several other ideal assumptions: a fully elastic collision – the velocity does not change before and after the collision; No loss of energy – no resistance; There is no time difference in motion – collision time is not counted. Under these assumptions, the movement of the two balls becomes very simple:

In the vertical direction: A and B balls are reciprocating free fall motion and their reverse motion; Because the energy loss is not counted, the two balls will definitely be able to bounce back to their original height.

Horizontally: b stationary; A moves in a straight line towards B.

Because the motion in the vertical direction is exactly the same, the balls A and B are "relatively stationary" in this direction. Therefore, if we take ball B as the frame of reference, then ball A is left with only horizontal motion: a uniform linear motion towards b.

So, no matter how high the horizontal initial velocity of the ball A is – as long as it is not zero – the ball A will always hit the ball B.

4/3。Let the time from the first to a time be t1, the horizontal velocity is v1, the second time is t2, the horizontal velocity is v2, and the ball falls from h to the ground time t, then there is t2=t, because the collision is a completely elastic collision, so the time for the ball to rise to h height after collision is also t, then there is t1=3t, then there is v1t1=v2t2, v2=3v1. Let the time from the first stroke to the ball wipe t3, and the second time from the stroke to the ball to the ball to wipe the net t4, then v1t3=v2t4---1).

t3 = 2 * root number (2gh) - root number (2g(h-h))-2), t4 = root number (2g(h-h)))-3). From (1), (2), (3), h h=4 3.

The time for the two balls to move flat is equal.

The ratio of horizontal range is s1:s2=1:3

The ratio of the initial velocity is v1:v2=1:3

The first ball (b---a) is thrown diagonally, h-h after horizontal distance = s'1+s'2=2s1

h= ; h-h=

and s1=v1*t1

s'1=v1*t2

s'2=v2*t2

v1*t2+v2*t2=2*(v1*t1)∴t1=2t2

h=4(h-h）

The solution is h:h=4:3

Hope it helps.

(1) The X direction and Y direction shown in the figure represent the X direction in the horizontal direction

2) the time interval between the balls passing through adjacent positions is equal; h1= h=gt^2/2 t1=

3) Find the magnitude of the initial velocity of the ball to do the flat throwing motion vo, vot1=x vo=

4) When the ball passes through the two points b and c, the velocity in the vertical direction is vy1 and vy2 respectively

As can be seen from the title, vy1 = gt1 = vy2 = g 2t1 = 1m s

5) If you forget to record the position of the thrown point a in the experiment, only the other three points b c d are marked

Then it can be obtained from the graph, ycd=5l, ybc=3l, xbc=xcd=2l, since the vertical motion is a free fall motion, so ycd-ybc=gt 2 can find the time interval t= (2l g).

Then the horizontal direction is divided into linear motion with uniform velocity, x=vot vo=x t=2l t= (2gl).

(1).x.

2）。Equal. The horizontal velocity of the horizontal movement of the flat toss motion is the same, and the time taken for the same displacement (two grids) is the same.

1/2gt*t=

t=(3)v=x/t=

4b point velocity: v(b)*v(b)-0*0=2*g*v(b)=

Point C velocity: v(c)*v(c)-0*0=2*g*5*v(c)=1m s(5).

The motion of the ball can be broken down into two directions: horizontal and vertical;

Move horizontally at a constant speed, and move in free fall vertically until you fall down the slope;

That is, when the ball hits the slope.

1. The distance traveled in the horizontal direction.

2. The distance from the vertical direction.

3. With the slope.

Just form a right triangle, let the motion time be: ttg = ( g t t) (v t);

After obtaining t, the horizontal velocity does not change = v;

Vertical velocity = gt;

The magnitude and direction of the combined velocity are no longer a problem.

When falling to point O, the vertical displacement y horizontal displacement x=tanay=(1 2)gt 2

x=vo*t

tana=(1/2)gt^2/(vo*t)=gt/(2vo)t=2vo*(tana)/g

vy=gt=2vo*tana

The speed sought is.

v = root number (vx 2 + vy 2) = root number [vo 2 + (2votana) 2] = vo * root number [1 + 4 (tana) 2].

Find the horizontal and vertical velocity at 0 point, respectively.

If you don't know, let's set it up.

Let the vertical velocity vy reach point O, and the time taken is t

Then the vertical displacement h= gt

Horizontal displacement x=h tan =vo·t

t=2votanα/g

vy=gt=2votanα

o point velocity v = vo + vy

The maximum distance is the most prominent part of the curve you draw on the diagram, and the velocity of the ball at the initial position is divided into the vertical and upward direction along the inclined plane, and when the upward partial velocity is reduced to 0, it reaches the maximum distance of the ball from the inclined plane.

The first question is that the distance is the largest when you land, you can find the distance between the highest point and the ground according to the angle and then calculate the time, multiply by v This is the maximum distance!

1 2gt 2=r, xoc=vt, xbc=xoc-xob, it is obtained, xbc = root number (2r g) v-r, and the root number (2r g) v is greater than r

Let the BC distance be l, and the object will move in a uniform circular motion around the sphere at a speed of v2, then:

vt=r+l；

1/2gt^2=r;

m（v2）^2)/r=mg;

v>v2;

The velocity v is greater than the root number (gr), and the BC distance is v*root number (2r g).

Find the landing time first, and then find the distance.

The analysis is wrong, it takes longer for the flying knife to hit point P, and it takes less time to hit point M h=>t=(2h g) (1 2)==>

h(m)=(2h/g)^(1/2)

h(p)=(2*3h/g)^(1/2)=(3)^(1/2)*h(m)>h(m)

Because point m is the farthest from point o among the three points, and the gravitational formula is the same, it takes a long time.

Answer: C flat throwing motion is only affected by gravity, the acceleration is unchanged, it is a uniform velocity curve motion of A=G, AB is wrong, C is correct. Since the velocity direction of the object in flat throwing motion when it lands is the combined velocity from the horizontal direction to the vertical direction, it must not be vertically downward, d is wrong.

a.False, the flat throwing motion is a uniform acceleration motion.

Acceleration is. g] b, false, x=vt, and .

Time-related. c, right, 1 2gt

HD, wrong, the direction of speed is constantly changing, however.

The direction of acceleration is always straight down

Of course, choose D! Uniform acceleration in the vertical direction and uniform speed in the horizontal direction, so the time is only determined by height. (1) False.

The acceleration is always g, and the direction is straight downward, so (3) is wrong, choose d: (2) and (4).

d is determined by h=1 2*gt*t, and the time is determined by height. The initial velocity in the horizontal direction does not change, which determines the range. The acceleration is always equal to g, unchanged. The velocity varies due to the presence of acceleration. The jerk does not change, and it is a uniform variable speed.

d(1) in The flight time is independent of speed.

3) The acceleration in the medium is unchanged.

The DA time is determined by the height, since 1 2 GT = H

b is correct c acceleration is g constant d is correct.